Best prof in the world !!! U have no idea of how much your videos has helped me in my life , prof ! I am in one of the top universities just by following your videos and studying everything in it !!!
I’m an incoming 2nd year Civil Engineering student and Diff equation is one of our subject. Before our school starts, I will definitely watch your diff eq. playlist (beginning to the end) And I will also watch your calc 1-3 playlist to refresh what we had studied last s.y. and also add knowledge that we haven’t tackled. Thank you for sharing your wisdom. You’re a huge part of my journey until I graduate and pass the licensure exam. Continue to be a blessing. THANK YOU PROFESSOR!
This greatly increased my understanding of kinematics. I always knew that vel is the first deriv of position and accel is the second deriv but did not have the necessary mathematical skills to apply this concept until now. Thank you so much.
I'm going to go through your whole differential equations playlist because I find it wonderful for my learning. Lots have been happening so I've lost my studying motivation, but your videos are helping me get on my feet again and I'm super thankful for that. Great and detailed explanations for every step and your calm teaching approach is just what I really needed. Thank you so much. =)
I have a chance in my physics major because of you. The professors at my university dont even teach, they just assign homework and test with less than 10 percent class time focused on instruction. Thank you.
Wish I come across these videos when I was taking D.E. These are awesome. On the truck problem, I did a slight variation. Since I knew I was going to have my first constant times t squared, I didn't solve for t, I left it as C * t = -88 and sub that in for C * t in the C * t squared so I didn't have to square 88, I would be dividing by 2 instead. Thinking hard about getting my Masters and wanted to brush up. Thanks for these.
I found this channel just about a month ago, and I’ve found everything you post really really helpful. Subscribed and liked, and also just love your personality! You’re a great teacher and a great person.
This was the only video of yours I found difficult to follow because a lot of it wasn't intuitive for me. But it all made a lot more sense in your summary at the end and after reviewing my notes. Thank you a lot.
Thank u Professor.I study Mechanical Engineering in Eskişehir,Turkey.This year I am taking course of Differential Equations and watching your videos.I'll probably prepare the exam with your videos :)
Hey, @Professor Leonard, can you make a movies about Number Theory, Linear algebra and Combinatorical Optimization? Otherwise you're the best professor I know! :)
You might find it strange but thhat’s the truth I am from a very far away place from Saudi Arabia but I really enjoy your work. What I only wish is to make writing a bit clearer. 👏👏👏👏
On the last problem if you divide the gravity of Earth by the gravity of "Planet", so (-32 ft/s / -10 ft/s) and multiply it by the maximum height achieved on Earth (144 ft). You get 460.8 ft. Just another way of getting the answer for those who are interested.
Professor, in the truck problem, if we have the deceleration to be negative doesn't it mean that our acceleration is positive, so the truck continues to move instead of stopping at a point. So we actually found out the acceleration.
I think in the question at part 38:20 it doesn't matter if you threw it up or down at 10 m/s because if you threw it up at 10 m/s it would pass the starting point at 10 m/s downwards anyway. I checked using both values and got the same answer. love physics, although it's a very idealised scenario aha
fun fact for the very last exemple you can directly do a cross product to find the height: Acceleration on the "planet" is 3.2 less than on Earth so the projectile will be 3.2 less affected by it so it will go 3.2 higher so 144*3.2 is 460,8
@@Alexmar345 an example of this is "s=ut+0.5at^2" which is an integral of "u=v+at" where u is initial velocity, v is final velocity, s is displacement, a is acceleration and t is time...
I find sketching a little diagram for these types of mechanics problems helps to visualize and think through much better what is physically happening. In my engineering kinematics class we would actually get 1 point just for having a diagram, even if we got the rest of the question wrong.
if u throw it up or down, it accelerates towards the ground. so whether u assume acceleration is positive or negative, adjust the rest of quantities to follow correct physics.
Regarding the watermelon example ,it`s completely irrelevant if we throw it up with the positive 10 m/s or throw it down with the negative 10m/s because once the watermelon levels with the initial position where we threw it from,the velocity will be negative 10m/s anyway.The displacement is just the same in either case.The watermelon just retraces its path between the positve 10 m/s velocity and the negative 10m/s velocity.The displacement has nothing to do with the retracing of the path.It`s just the difference between the initial and the final position.I hope this can help .
HELP please. at 1:15 you insert 32t for initial velocity, but you don't use it to calculate. im sure its obvious and i've just been staring at math for too long. if you see this, anyone, would you explain why he then calculates t with 'just' 144=16t^2 and doesn't use t, after inserting it in the line above...is it that 32t = 0?
On Ex5 the signs are weird, if you are throwing it down with velocity -10, it's the same as throwing it up at +10. It threw me off because i tried different speeds at which the W.M hits ground for 2 example: 1) how u first wrote it (thrown up) 2) the worked example (thrown down) Both gave the same answer for the height of the building, took me a while to figure out it was the same question xD Ty for excellent video regardless, been loving the Diff Eq series ^^
Faster refers to a magnitude. The magnitude of velocity is increasing, in the negative direction. Negative does not always refer to a decrease in magnitude
In the case of the truck sliding to a stop problem, isn't finding the solution easier when we also use the fact that the end velocity (V) equals zero, which means we have all our variables besides A (a constant and function of/ultimately determined by end T) and T (the only actual value we need to find)? With X0 = 0, X = 176, V0 = 88, and V = 0, V(T) = A*T + 88, and X(T) = 0.5*A*T^2 + 88T. With V = 0, and knowing all other values, plug V into V(T) with 0 = A*T + 88, get A = -88/T, then plug that into the equation X(T), with 176 = -44T + 88T = (-44+88)T = 44T. Solve for T, getting T = 4 seconds, and then solve for A using A = -88/T, and get A = -22 m/s^(-2). Since it isn't systematic, this would probably only save time in this particular instance. Or it might rely on assumptions that can't always be made. But I would like to know if I did something illogical (and if so, what it was and why).
In the truck problem it would be easier for us to plug a=(-88ft/s)/t into the position equation and solve for t, much smaller numbers and no dealing with fractions. Still good video and learned a lot from it, thanks😃.
if u assume a directional quantity to be positive while moving towards the ground, u assume any in the direction of it to be positive, and any in opposite direction of it to be negative. whichever u assume, you will get the same result.
I solved the watermelon problem assuming we throw it up and got the same answer of 178.57 meters (1250/7 mts to be more precise). So, is the height of the building the same regardless of the direction of our throw? Does that mean that the height depends only on the final velocity? (Other data I got is it reaches maximum height at t=(50/49) seconds and it hits the ground in t=(350/49) seconds). Great video btw :)
Hi Professor Leonard. First of all thanks for making these videos they are very useful to me. Secondly what camera are you recording with? I need something that gives me the same quality for work. Thank you.
At 29:53 t=10.2 seconds where velocity =0. One could argue the velocity is 0 at the max height of flight and after it hits the ground. What is wrong with the latter assumption? It is confusing because just as it hits its non zero but instantly afterward it is 0. If there are two choices for t=0, how do you know which one to use?
Whew! I'm tired after working through those with you. I'm still confused as to the general forms these problems can take. I guess I'll have to review this video (and maybe the one before this one).
PSA: in the car skid question (176 foot slide) you can solve for time first (sub c1t = -88 in the equation for x) and then do acceleration. Nicer numbers.
33:52 We’ll you didn’t even have to calculate that 20.4 seconds bc you that if it reaches its maximum height in 10.2 seconds, bc of time dilation, it will return to the position of 0 displacement in double the time it takes to reach its maximum height
First of all, thanks for the awesome videos! I think you have the watermelon problem (ending around 43:43) mixed up however. You got 178.4m, which is the height I ended up with when I did it assuming the initial velocity was POSITIVE 10m/s (throwing the watermelon UP, as you had the problem originally worded). If the watermelon is thrown down at 10m/s I get 76.4m (when I plug in t = 5.1 in your position function I end up with 76.4 for C as well).
Hey there! Thanks for watching. Both the 60 and the 10 should be negative (towards the ground). Check to see if you're getting a negative t or a positive t. It should be positive.
Ah I see now. When I first saw that you got 178.4m I thought it had to be wrong, since I got the same value assuming we were throwing the watermelon up (and after checking my work a bunch of times). But now that I think about it, given that i assumed the initial velocity was 10, and given that g is almost -10, the watermelon would be returning to the initial thrown height moving at almost exactly at -10 (it ends up being -9.99 m/s if you take g to be -9.8), like you have it set up in the problem. This would then of course give us the same final answer of 178.4. As far as 76.4 goes, I have no idea how I arrived at that now. Pretty sure I entered something into the calculator wrong. Anyway, thanks again!
Hi Prof. In your video, Differential Equations 8 Example 3, why is acceleration a= -32 taken to be negative yet it is pointing towards the ground, that is, in the direction of gravity?
wouldn't initial velocity be affected by acceleration? since earth and "planet" have different acceleration, how could we just use initial velocity from Earth to substitute initial velocity on "planet"
I have confused about one of the questions at 1Hr:05min and 13 sec, he said that velocity not going to change due to acceleration is constant, but velocity changing because there are acceleration exist. I am a little bit confused with that statement. Anyone can explain?
Honest Question... IF I completed the Physics Engineering Series should we skip this video? Skipping through the video, i did this stuff for 3 semesters xD
Because you take the up direction as positive, and acceleration is constantly pulling downwards. On the breaks example, acceleration goes in the direction contrary to the movement you are considering as positive
Because acceleration due to gravity points downwards if you think in terms of the Cartesian coordinate system where positive y points up and negative y points down
The coordinate system you choose is arbitrary. In this case, Professor decided to have "up" be positive direction and "down" be negative direction. So the signs of acceleration, velocity and position correspond to that. You could also choose to have down be positive and up be negative, but in that case you'll be dealing with negative values for position. Either way you do it, you'll get the right answer. The important thing to remember is to stick to the coordinate system you chose for your entire problem.
You totally ignored one thing, the sign rule, what it represents. We are not just dealing with speed in example 2, we're also dealing with velocity, which has importance for sign. Here's the thing. We took upward to be positive, and gravitational acceleration works downwards, so it's negative 9.8, height is considered from ground to upward, so it's positive. The result in 2nd example, V(t) = -35ft/s indicates that the magnitude is 35ft/s and the direction is downward.
Best prof in the world !!! U have no idea of how much your videos has helped me in my life , prof ! I am in one of the top universities just by following your videos and studying everything in it !!!
I’m an incoming 2nd year Civil Engineering student and Diff equation is one of our subject. Before our school starts, I will definitely watch your diff eq. playlist (beginning to the end) And I will also watch your calc 1-3 playlist to refresh what we had studied last s.y. and also add knowledge that we haven’t tackled. Thank you for sharing your wisdom. You’re a huge part of my journey until I graduate and pass the licensure exam. Continue to be a blessing. THANK YOU PROFESSOR!
Aaah. The one hour long videos are back! Love it.
sometimes i miss 3 hours long videos, just kiddin but leonard made it affordable
This greatly increased my understanding of kinematics. I always knew that vel is the first deriv of position and accel is the second deriv but did not have the necessary mathematical skills to apply this concept until now. Thank you so much.
Sir you are really the best.......Thank you for helping all of us in our studies
This man is a gift from the math gods...a national treasure. May he live long and prosper.
I'm going to go through your whole differential equations playlist because I find it wonderful for my learning. Lots have been happening so I've lost my studying motivation, but your videos are helping me get on my feet again and I'm super thankful for that. Great and detailed explanations for every step and your calm teaching approach is just what I really needed. Thank you so much. =)
Practically, I would just derive the kinematic formuals and use those, but this thoroughly showed basically the same thing each time. Loved the video!
I have a chance in my physics major because of you. The professors at my university dont even teach, they just assign homework and test with less than 10 percent class time focused on instruction. Thank you.
Wish I come across these videos when I was taking D.E. These are awesome. On the truck problem, I did a slight variation. Since I knew I was going to have my first constant times t squared, I didn't solve for t, I left it as C * t = -88 and sub that in for C * t in the C * t squared so I didn't have to square 88, I would be dividing by 2 instead. Thinking hard about getting my Masters and wanted to brush up. Thanks for these.
the best prof I have ever seen in my life. i hope that you will not hesitate to help us. as mathematicians, we say we love you.
Great video! Glad to finally find good differantial equations playlist :D
1:25:35 Yes....!!! it did help, you explained it so well thank you , please don't stop making theses DE videos, and have nice day proffeser.
This is an awesome review. Took differential eqs about 8 years ago. My professor sucked so bad! Now its all starting to make sense. Thanks!
I found this channel just about a month ago, and I’ve found everything you post really really helpful. Subscribed and liked, and also just love your personality! You’re a great teacher and a great person.
This was the only video of yours I found difficult to follow because a lot of it wasn't intuitive for me. But it all made a lot more sense in your summary at the end and after reviewing my notes. Thank you a lot.
And Merry Christmas!
And Merry Christmas to you!!
Thank u Professor.I study Mechanical Engineering in Eskişehir,Turkey.This year I am taking course of Differential Equations and watching your videos.I'll probably prepare the exam with your videos :)
YOU ARE THE BEST! You make my life much easier. Please keep making videos.
I wish you were my professor for the upcoming fall semester on differential equations. Your videos are totally helping me get prepared. Thank you!!
PROFESSOR LEONARD IS THE GOAT
this was fantastic, I wish there were videos like this for linear algebra...and all my other courses.
Thanks so much for working these out. Those last few were a lot harder! I need to practise more that are like that.
Best lecturer like ever..... Keep teaching people how to love math :-)
Hey, @Professor Leonard, can you make a movies about Number Theory, Linear algebra and Combinatorical Optimization? Otherwise you're the best professor I know! :)
You might find it strange but thhat’s the truth I am from a very far away place from Saudi Arabia but I really enjoy your work.
What I only wish is to make writing a bit clearer.
👏👏👏👏
You're an awesome human being , Professor!
An excellent teacher professor Leonard. Thanks!
On the last problem if you divide the gravity of Earth by the gravity of "Planet", so (-32 ft/s / -10 ft/s) and multiply it by the maximum height achieved on Earth (144 ft). You get 460.8 ft. Just another way of getting the answer for those who are interested.
Please include differential equations application to electrical circuits pleaseeeee
To be honest, after this course everything should be clear.
This is actually better than using kinematic equations in my opinion.
That's because Physics makes a lot more sense when Calculus is involved.
Very well explained. I really liked it.
Continue your good work.
Professor, in the truck problem, if we have the deceleration to be negative doesn't it mean that our acceleration is positive, so the truck continues to move instead of stopping at a point. So we actually found out the acceleration.
I think in the question at part 38:20 it doesn't matter if you threw it up or down at 10 m/s because if you threw it up at 10 m/s it would pass the starting point at 10 m/s downwards anyway. I checked using both values and got the same answer. love physics, although it's a very idealised scenario aha
fun fact for the very last exemple you can directly do a cross product to find the height: Acceleration on the "planet" is 3.2 less than on Earth so the projectile will be 3.2 less affected by it so it will go 3.2 higher so 144*3.2 is 460,8
Thanks! This really helps out with A level maths and physics
(It’s also easier to use SUVAT equations but this is a video on calculus after all...)
DANK BOI suvat?
@@Alexmar345 an example of this is "s=ut+0.5at^2" which is an integral of "u=v+at" where u is initial velocity, v is final velocity, s is displacement, a is acceleration and t is time...
@@michaelhunt2222 we ARE using suvat equation here. s(t)=1/2 a t^2+ v(0)t+s(0)
I find sketching a little diagram for these types of mechanics problems
helps to visualize and think through much better what is physically
happening. In my engineering kinematics class we would actually get 1
point just for having a diagram, even if we got the rest of the question
wrong.
@3:35 😂 set up for some hilarious math stand up comedy
hi , in watermelon example @39:20 when body is moving downwards why acceleration is taken negative 9.8?
if u throw it up or down, it accelerates towards the ground.
so whether u assume acceleration is positive or negative, adjust the rest of quantities to follow correct physics.
yeahh long videos thats what I want
I prefer the shorter ones actually. :(
@@taekwondotime
The longer videos do have breaks here and there though, so you could think of them as several short videos thrown together. :)
You are the best!!! (Consider adding a diagrams and keep the units in the equations.) (Please don't say that time and velocity are signless. LOL)
"Time is the one dimension you can't go back on...its called regrets...no regrets" ~Professor leonard
These examples are very helpful. Thank you!
legit super amazing prof! you nailed it! thanks a lot :D
Thank you man for your time! THANK YOU!
A lot of examples really helps ! thanks
Regarding the watermelon example ,it`s completely irrelevant if we throw it up with the positive 10 m/s or throw it down with the negative 10m/s because once the watermelon levels with the initial position where we threw it from,the velocity will be negative 10m/s anyway.The displacement is just the same in either case.The watermelon just retraces its path between the positve 10 m/s velocity and the negative 10m/s velocity.The displacement has nothing to do with the retracing of the path.It`s just the difference between the initial and the final position.I hope this can help .
13:47 "No regrats! Not even a letter..." is that a We're the Millers reference? :D
HELP please. at 1:15 you insert 32t for initial velocity, but you don't use it to calculate. im sure its obvious and i've just been staring at math for too long. if you see this, anyone, would you explain why he then calculates t with 'just' 144=16t^2 and doesn't use t, after inserting it in the line above...is it that 32t = 0?
On Ex5 the signs are weird, if you are throwing it down with velocity -10, it's the same as throwing it up at +10.
It threw me off because i tried different speeds at which the W.M hits ground for 2 example:
1) how u first wrote it (thrown up)
2) the worked example (thrown down)
Both gave the same answer for the height of the building, took me a while to figure out it was the same question xD
Ty for excellent video regardless, been loving the Diff Eq series ^^
yeah me too😁
It's called "REGRETS". I LOVE YOU :,,,)
10:00 why a is a negative it's getting faster shouldn't the acceleration be positive?
Faster refers to a magnitude. The magnitude of velocity is increasing, in the negative direction. Negative does not always refer to a decrease in magnitude
it was a great opportunity for me to learn from lifting experts from all industries. including my mentor Mr Utibe Akpanudo
In the case of the truck sliding to a stop problem, isn't finding the solution easier when we also use the fact that the end velocity (V) equals zero, which means we have all our variables besides A (a constant and function of/ultimately determined by end T) and T (the only actual value we need to find)?
With X0 = 0, X = 176, V0 = 88, and V = 0,
V(T) = A*T + 88,
and
X(T) = 0.5*A*T^2 + 88T.
With V = 0, and knowing all other values, plug V into V(T) with
0 = A*T + 88,
get A = -88/T, then plug that into the equation X(T), with
176 = -44T + 88T = (-44+88)T = 44T.
Solve for T, getting T = 4 seconds, and then solve for A using A = -88/T, and get A = -22 m/s^(-2).
Since it isn't systematic, this would probably only save time in this particular instance. Or it might rely on assumptions that can't always be made. But I would like to know if I did something illogical (and if so, what it was and why).
Sorry but how is this any different from the way he did it?
You went in very depth....thanks for that😊
On this day I fell in love The day this video was published 2018101
It was great watching... thanks for the help..
In the truck problem it would be easier for us to plug a=(-88ft/s)/t into the position equation and solve for t, much smaller numbers and no dealing with fractions. Still good video and learned a lot from it, thanks😃.
Why is acceleration -ve in 12:30? If I'm right, when we drop the ball, g speeds up the ball not decrease so why the -ve sign??
if u assume a directional quantity to be positive while moving towards the ground, u assume any in the direction of it to be positive, and any in opposite direction of it to be negative.
whichever u assume, you will get the same result.
I solved the watermelon problem assuming we throw it up and got the same answer of 178.57 meters (1250/7 mts to be more precise). So, is the height of the building the same regardless of the direction of our throw? Does that mean that the height depends only on the final velocity? (Other data I got is it reaches maximum height at t=(50/49) seconds and it hits the ground in t=(350/49) seconds). Great video btw :)
Thanks so much professor
Thank you! I did find this video helpful.
Hi Professor Leonard.
First of all thanks for making these videos they are very useful to me.
Secondly what camera are you recording with? I need something that gives me the same quality for work. Thank you.
At 29:53 t=10.2 seconds where velocity =0. One could argue the velocity is 0 at the max height of flight and after it hits the ground. What is wrong with the latter assumption? It is confusing because just as it hits its non zero but instantly afterward it is 0. If there are two choices for t=0, how do you know which one to use?
Professor, what textbook do you recommend to supplement these videos?
Whew! I'm tired after working through those with you. I'm still confused as to the general forms these problems can take. I guess I'll have to review this video (and maybe the one before this one).
God be pleased with you,inşAllah,Amen🤲
Without having to do the quadratic, there is the kinematic equation of V^2=V^2+2ad, then solve for time with v=v+at.
we all know that. But using diferential equations is quite nice
Does he use a particular textbook to supplement these lectures or are they all based on the notes given in each video?
Absolute legend
PSA: in the car skid question (176 foot slide) you can solve for time first (sub c1t = -88 in the equation for x) and then do acceleration. Nicer numbers.
I love you man ❤
From the apex (v=0) to the terminal velocity (v=60) 60 = 0 + 9.8t t=6.12
awesome 👏
33:52 We’ll you didn’t even have to calculate that 20.4 seconds bc you that if it reaches its maximum height in 10.2 seconds, bc of time dilation, it will return to the position of 0 displacement in double the time it takes to reach its maximum height
that was helpful, i appreciate it :)
First of all, thanks for the awesome videos!
I think you have the watermelon problem (ending around 43:43) mixed up however. You got 178.4m, which is the height I ended up with when I did it assuming the initial velocity was POSITIVE 10m/s (throwing the watermelon UP, as you had the problem originally worded). If the watermelon is thrown down at 10m/s I get 76.4m (when I plug in t = 5.1 in your position function I end up with 76.4 for C as well).
Hey there! Thanks for watching. Both the 60 and the 10 should be negative (towards the ground). Check to see if you're getting a negative t or a positive t. It should be positive.
Ah I see now. When I first saw that you got 178.4m I thought it had to be wrong, since I got the same value assuming we were throwing the watermelon up (and after checking my work a bunch of times). But now that I think about it, given that i assumed the initial velocity was 10, and given that g is almost -10, the watermelon would be returning to the initial thrown height moving at almost exactly at -10 (it ends up being -9.99 m/s if you take g to be -9.8), like you have it set up in the problem. This would then of course give us the same final answer of 178.4.
As far as 76.4 goes, I have no idea how I arrived at that now. Pretty sure I entered something into the calculator wrong. Anyway, thanks again!
Superman believes in gravity! Unreal.
Hi Prof. In your video, Differential Equations 8 Example 3, why is acceleration a= -32 taken to be negative yet it is pointing towards the ground, that is, in the direction of gravity?
Yes bro even in problem 2 🙄
Thank you sir!! Going to be helping me out with my physics kinematics!
wouldn't initial velocity be affected by acceleration? since earth and "planet" have different acceleration, how could we just use initial velocity from Earth to substitute initial velocity on "planet"
It was very helpful :)
Are there any examples involving non constant acceleration?
Aren’t these problems just simply vertical projectile motion questions?
I have confused about one of the questions at 1Hr:05min and 13 sec, he said that velocity not going to change due to acceleration is constant, but velocity changing because there are acceleration exist. I am a little bit confused with that statement. Anyone can explain?
Just curious, do you "play" with Matlab, SageMath, and so on?
An hour and a half, whoa!
What if you use the pre-defined equations of motion such as:
v=u+gt
h=ut+1/2gt^2
v^2=u^2+2as
?
Yeah, but you need to understand from where these formulas come from ...
@@kesjerdna which I do
Time is only thing that can't go negative. Time is the only Dimension you can't go backwards on, it's called regrets.
Everything in Physics 1 just clicked. :O!
34:30 Does anyone know how he got 20.61s? Thanks!
thanks
I, too, love paintball.
Last few examples might be nice
I do not think the example question at 38' is correct if you re-write the question "thrown down" instead of "thrown up".
v=+10m/s and g=+9.8m/s^2
Hi, according to the book it's correct. He might be going based on the book.
"No regrets." lol
Honest Question... IF I completed the Physics Engineering Series should we skip this video? Skipping through the video, i did this stuff for 3 semesters xD
i don't understand why acceleration is taken as negative when throwing something from above to the ground , can anyone help
Because you take the up direction as positive, and acceleration is constantly pulling downwards. On the breaks example, acceleration goes in the direction contrary to the movement you are considering as positive
any hint ??
Because acceleration due to gravity points downwards if you think in terms of the Cartesian coordinate system where positive y points up and negative y points down
The coordinate system you choose is arbitrary. In this case, Professor decided to have "up" be positive direction and "down" be negative direction. So the signs of acceleration, velocity and position correspond to that. You could also choose to have down be positive and up be negative, but in that case you'll be dealing with negative values for position. Either way you do it, you'll get the right answer. The important thing to remember is to stick to the coordinate system you chose for your entire problem.
i had my silingshot out :)
You totally ignored one thing, the sign rule, what it represents. We are not just dealing with speed in example 2, we're also dealing with velocity, which has importance for sign. Here's the thing.
We took upward to be positive, and gravitational acceleration works downwards, so it's negative 9.8, height is considered from ground to upward, so it's positive. The result in 2nd example, V(t) = -35ft/s indicates that the magnitude is 35ft/s and the direction is downward.
the best