My phys teacher went over this but he talks so fast and just chicken scratches examples and notes for us and I cannot keep up like I remember him going through this but he moved so fast I just got lost in class and just gave up on trying to listen anymore and kept trying to remember the last example he erased to copy into my notes. So nice to have someone slow down and explain things in a relaxed manner instead of like we’re gonna go as fast as possible hope u can keep up here’s the hw any questions? Ask question .... where are the notes u need to use these equations look at my notes and half the equations had errors in them. Thank you it’s not so hard when the information is being shot at you at 299792458m/s
Same happened with me during my class 12th days physics concepts were taught so fast as the focus was on completing syllabus i guess ...its not cool... i couldn't grasp most of the concepts and the fact that i did not take tutions made it even harder .
I was seriously considering changing my major so I could avoid taking college physics but your videos are saving me! You make it so simple and easy to understand. With Covid, I can't get any help and my professors have basically checked out, but I was able to do my homework with only, like, two crying breaks.
the Bonus Question # 01 can actually solved with this equation = Maximum Height plus the given displacement of the y component. H = 3.72m plus 15 m = 18.7 m...
Physics ninja why do your numbers not add up, I'm confused because I'm using my calculator for calculating the final velocity and when I solve for 8.5-9.8(2.82) its equal to 19.1m/s
This might be a dumb question but wouldn’t the final velocity of the object be negative because it’s pointing down? I know the negative goes away when you square the terms but shouldn’t the resulting vector be negative to properly denote direction?
For part a are you factoring in the time the object takes while it is above the launch point? I solved for the time it takes to reach the highest point in its path, multiplied by 2, and then found the velocity at that time, which I used to find the time for the object to move 15m. I ended up with 4.52 seconds
Hold on….. you said that the velocity at the max height is not zero as it has some velocity in the horizontal direction . I used that formula and I still got the same answer. You can still use velocity at max = zero. Technically speaking, the particle has stopped moving at the highest point. We know that vox = vx because the velocity is constant in the horizontal direction which is fine. But remember we are asked about the y direction so I’d use the velocity in the y direction zero as the particle has stopped moving momentarily.
@@abridgetool That's because you used a different Gravitational Constant. You used g = 9.81, and Physics Ninja used 9.8. If you go through with it to the end, you will get that the final answer (resultant) is correct.
if the given value of sine 25 (-0.13235175009) would remain in radian ? will it be same (horizontal Range) as the answer for sine 30 degree ? If it isn't. Would you please inform me how can i get rid from this .((in c and java there is a function for radian instead of degree , and that's why when i'm going to solve my problem that become confrontational with my output..))
I computed the max height before I computed the air-time by Integrating and adding a Constant. I was just checking if my method worked universally and I think I have gotten this down now. Nonetheless, it is always good to verify what I know is correct. Thanks for this video.
I cannot find any examples of what to do if you are NOT given the angle. What if you are given a target range, the height of the cliff, and an initial speed. How would you calculate the angel needed?
yes, range is given, but nothing else. I've tried doing it using equation substitution, but I'm having a hard time cleaning up and simplifying the trig @@PhysicsNinja
Hi. I was wondering how to solve this without being given the height of the cliff. My teacher didn’t give the height of the cliff which made this so much harder
I have this question for a while now. Given the elevation, the range, and the initial velocity, how do you find the angle by which the projectile was launched
i have the same question, though i have progress. first is you need to need to get the distance of the projectile from initial position until the vertical velocity of the projectile is zero. v_x*v_y/g where g is gravity, v_x is horizontal velocity, and v_y is vertical velocity. second, you need to get the time until the projectile touches the ground, starting from the time where vertical velocity is zero (from step 1), using the equation t = √(2d/g) where t is time, d is displacement, g is gravity. the value of d is the highest height the projectile had (including elevation) using the equation d = v_y*t+0.5*at^2+h where, t is the time the projectile is at it's highest value, h is the elevation relative to the height of the ground where the projectile lands. the value of t is equal to vsin(o)/g basically you get this equation (not simplified) o = angle range = (vcos(o)*vsin(o)/g)+✓(2(vsin(o)*(vsin(o)/g)+(1/2)*g*(vsin(o)/g)^2)/g) third, you rearrange the equation, where the equation equals the angle, which is my problem.
Kariuki Ke you need to find the time when its at the max height first, this is when the vertical component of the velocity is 0. Once you have this time plug back into y equation and find the height.
Homie your numbers dont add up?? the final velocity in the y axis is calculated to be 19.11 m/s vfy ^2 = viy^2 + 2ady => vfy = √viy^2 + 2ad = √(20sine25)^2+ (2x15x9.8) = 19.11 m/s and then you use vfy to calculate the time using the eqaution vfy-viy/9.8 = 1.087 seconds edit : the final velocity you got is the final velocity XD, such a coincidence. The time is wrong tho-
Then you need to work backwards. If you know the range of the projective R and the time you can find the velocity in the x-direction. If you know the time and the height you can use this information to find the initial velocity in the vertical direction. Once you know both components of the initial velocity solve for the launch angle using some trigonometry.
`wouldn't you multiply time by two when finding how much distance travelled because you found the time for the vertical which is only half the parabola, so the time would be twice as much when finding the horizontal distance which is the whole parabola?
That's only true if you launch something from the ground and it land on the same flat ground. In that case the problem is completely symmetric. In the case illustrated above, we only have some of the parabola on the left side compared to the right hand side.
My phys teacher went over this but he talks so fast and just chicken scratches examples and notes for us and I cannot keep up like I remember him going through this but he moved so fast I just got lost in class and just gave up on trying to listen anymore and kept trying to remember the last example he erased to copy into my notes. So nice to have someone slow down and explain things in a relaxed manner instead of like we’re gonna go as fast as possible hope u can keep up here’s the hw any questions? Ask question .... where are the notes u need to use these equations look at my notes and half the equations had errors in them. Thank you it’s not so hard when the information is being shot at you at 299792458m/s
Relatable
Same happened with me during my class 12th days physics concepts were taught so fast as the focus was on completing syllabus i guess ...its not cool... i couldn't grasp most of the concepts and the fact that i did not take tutions made it even harder .
I was seriously considering changing my major so I could avoid taking college physics but your videos are saving me! You make it so simple and easy to understand. With Covid, I can't get any help and my professors have basically checked out, but I was able to do my homework with only, like, two crying breaks.
I have online course on Udemy that you access from my website www.physicsninja.org. They cover most topics in first year physics.
Physics Ninja, good job! Thank you! I hope you are healthy and happy and making more videos because this is super helpful.
I was so nervous about my test! This helped so much, thank you!!!
Best of luck of your test! You got this.
the Bonus Question # 01 can actually solved with this equation = Maximum Height plus the given displacement of the y component. H = 3.72m plus 15 m = 18.7 m...
AP Physics Exam in two hours! This was the one thing I was confused about. Now I'm not! Thanks!
you deserve more subscribers sir. Thank you so much for yr help!
Thanks man you really helped me before my test 😂
kinda weird how a 13 minute youtube video explains the subject 10x better than my university professor I pay thousands to see
Thanks Tyler! Good luck with your physics class.
What if we are given only the distance traveled and not the initial velocity or time?
great vid but you gotta put the mic a bit closer, it's kind of hard to hear you.
exactly
turn up the volume idiot
for sure
Physics ninja why do your numbers not add up, I'm confused because I'm using my calculator for calculating the final velocity and when I solve for 8.5-9.8(2.82) its equal to 19.1m/s
Same here
yeah, he did it wrong-
This might be a dumb question but wouldn’t the final velocity of the object be negative because it’s pointing down? I know the negative goes away when you square the terms but shouldn’t the resulting vector be negative to properly denote direction?
You can decide which directions are positive or negative however I think I becomes negligible at that point and they are only interested in the speed
HA I just had the same question. The final velocity should DEFINITELY be negative if the initial velocities are positive.
Great video, thanks so much for the help!!!!
For part a are you factoring in the time the object takes while it is above the launch point? I solved for the time it takes to reach the highest point in its path, multiplied by 2, and then found the velocity at that time, which I used to find the time for the object to move 15m. I ended up with 4.52 seconds
You make this a lot easier
Where did you get the 20 that you multiplied with sine 25
Omg thank you so much, I was seriously considering just giving up until I saw this video. Thank you🙏🙏🙏
Hold on….. you said that the velocity at the max height is not zero as it has some velocity in the horizontal direction . I used that formula and I still got the same answer. You can still use velocity at max = zero. Technically speaking, the particle has stopped moving at the highest point. We know that vox = vx because the velocity is constant in the horizontal direction which is fine. But remember we are asked about the y direction so I’d use the velocity in the y direction zero as the particle has stopped moving momentarily.
Yes you can analyze the y component and set vy to 0. No problem there.
for the find the velocity question, i tried to solve it by finding the hmax first then using the formula Vf^2=Vi^2-2gh but it wasn't the same as yours
I am getting 19.11 m/s and it is not the same as the solved answer.
@@abridgetool That's because you used a different Gravitational Constant. You used g = 9.81, and Physics Ninja used 9.8. If you go through with it to the end, you will get that the final answer (resultant) is correct.
Me too...I get like 19.2
@@samisiddiqi5411 mf- its the same? the only difference is is 0.01
That was super helpful! Thank you
if the given value of sine 25 (-0.13235175009) would remain in radian ? will it be same (horizontal Range) as the answer for sine 30 degree ? If it isn't. Would you please inform me how can i get rid from this .((in c and java there is a function for radian instead of degree , and that's why when i'm going to solve my problem that become confrontational with my output..))
why don't you make 9.8 negative separate from the equations subtraction sign?
THANKYOU
Thank you Mr
So if i throw a ball with same angle and power from the ground it will land in the same place?
Thank you so much! Totally saved me :)
Hello i think there is an error for the time. Shouldn't the time formula be: Vy x 2/98? Which would give you 1.73 second
TopKevessi yea that’s what I got
when I plugged this info into a projectile calculator the answers weren’t the same but very close, what could be the reason for that?
why do I get an error of an unreal number when I plug in the values for the first equation
In the quadratic equation I got 2 negatives, my a was 4.9 b=9.17 c=0.78 meters. The time cannot be negative and I don't know where I went wrong.
a should probable be negative I’m guessing
A projectile is thrown from a height H. Find theta for maximum ground range? Pls solve
I computed the max height before I computed the air-time by Integrating and adding a Constant. I was just checking if my method worked universally and I think I have gotten this down now.
Nonetheless, it is always good to verify what I know is correct. Thanks for this video.
That’s just wrong method of doing it. So you might end up losing marks if the instructions says otherwise.
Why is it - 1/2 x g x t squared?
I cannot find any examples of what to do if you are NOT given the angle. What if you are given a target range, the height of the cliff, and an initial speed. How would you calculate the angel needed?
You will need to work backwards. You are most likely given something else, range, max height.
yes, range is given, but nothing else. I've tried doing it using equation substitution, but I'm having a hard time cleaning up and simplifying the trig @@PhysicsNinja
*Is it possible to solve this if Vo is also an unknown?*
how could the initial velocity be worked out?
Hi. I was wondering how to solve this without being given the height of the cliff. My teacher didn’t give the height of the cliff which made this so much harder
I am in class 9 and can solve these question
Which is hard for u
What does b stand for?
great job thanks
you just save me from hell! Thank you HAHHAHAHAHAH
Glad I could help!
wow!!! thank you !!!
Thank you so much for this video! It helps a lot when someone walks you step by step!
How do you find time if you only have theta, initial velocity, initial velocity of x, and gravity?
you will need to use the y equation and solve a quadratic equation for time. You should also be given the height of the cliff.
I have this question for a while now. Given the elevation, the range, and the initial velocity, how do you find the angle by which the projectile was launched
i have the same question, though i have progress.
first is you need to need to get the distance of the projectile from initial position until the vertical velocity of the projectile is zero. v_x*v_y/g where g is gravity, v_x is horizontal velocity, and v_y is vertical velocity.
second, you need to get the time until the projectile touches the ground, starting from the time where vertical velocity is zero (from step 1), using the equation t = √(2d/g) where t is time, d is displacement, g is gravity.
the value of d is the highest height the projectile had (including elevation) using the equation d = v_y*t+0.5*at^2+h where, t is the time the projectile is at it's highest value, h is the elevation relative to the height of the ground where the projectile lands.
the value of t is equal to vsin(o)/g
basically you get this equation (not simplified)
o = angle
range = (vcos(o)*vsin(o)/g)+✓(2(vsin(o)*(vsin(o)/g)+(1/2)*g*(vsin(o)/g)^2)/g)
third, you rearrange the equation, where the equation equals the angle, which is my problem.
@@panzerr4627 Yeah I'm also with you right there, you can't completely isolate the angle
What is Tup and Tdown??
But what if you want to find the height?
Kariuki Ke you need to find the time when its at the max height first, this is when the vertical component of the velocity is 0. Once you have this time plug back into y equation and find the height.
What happens in the case of the discriminant being negative? Surely there’s still a real solution to that physics problem
Homie your numbers dont add up?? the final velocity in the y axis is calculated to be 19.11 m/s
vfy ^2 = viy^2 + 2ady => vfy = √viy^2 + 2ad = √(20sine25)^2+ (2x15x9.8) = 19.11 m/s
and then you use vfy to calculate the time using the eqaution vfy-viy/9.8 = 1.087 seconds
edit : the final velocity you got is the final velocity XD, such a coincidence. The time is wrong tho-
I've tried solving the question countless times, I've been landing on the time of
t= 1.41sec.
what if you needed to solve for the launch angle???
Then you need to work backwards. If you know the range of the projective R and the time you can find the velocity in the x-direction. If you know the time and the height you can use this information to find the initial velocity in the vertical direction. Once you know both components of the initial velocity solve for the launch angle using some trigonometry.
Ahh, thanks Ninja. Figured it out! Dynamics is hard!!!
@@PhysicsNinja what if you only know the range, the elevation, and the initial velocity, how would you solve for the angle?
Hi, i was wondering for part 1 what g stands for and how to find it. Thanks
g is the acceleration due to gravity. Close to the earth g=9.8m/s2 (down)
The problem given to me was to find the maximum height. The only given in the problem is the velocity v and the angle theta. Idontevenknow.
Good video just you kinda sound like Saul Goodman.
`wouldn't you multiply time by two when finding how much distance travelled because you found the time for the vertical which is only half the parabola, so the time would be twice as much when finding the horizontal distance which is the whole parabola?
That's only true if you launch something from the ground and it land on the same flat ground. In that case the problem is completely symmetric. In the case illustrated above, we only have some of the parabola on the left side compared to the right hand side.
How fast do I need to be going to make sure my dog & I will die.
i get negative time and its is very small value
what if you needed to solve for time taken for peak height?
thats just -b/2a
@@vaneet7161 the x component of the vertex
closer to mic please. it sounds like your mic is in florida and youre speaking from berlin
what if the H is missing help huhu
Why isn't 9.8 negative
Helpful but audio is terrible!!! So took me a while to get it
We more volume profe,please...
could any one help me with a phisics problem please !!!!!!!!!
But I got -2.647 when I multiplied 20 x sine 25.
I'm confused 😔😔😔
your calculator is in radians. switch to degrees
Thank you so much
PHYSICS NINJA COULD BE BETTER==ZERO BABBLING INSTRUCTOR VIDEO++INSTEAD EASY 2 VIEW GEOMETRY 2 HELP SOLVE PUZZLES.
Where did you get the 20 that you multiplied with sine 25
Where did you get the 20 that you multiplied with sine 25
Where did you get the 20 that you multiplied with sine 25
Where did you get the 20 that you multiplied with sine 25
Where did you get the 20 that you multiplied with sine 25
thats the velocity initial
Where did you get the 20 that you multiplied with sine 25
Its the initial velocity of the object being thrown 20m/s
@@unauthordox8830 got it! Thanks bro