in last question, seperate into two parts ( ascend and fall), 2.55 sec for ascend, v=0+2(-9.18)(-132), s=vt/2, then u get t for fall: 5.39. then (5.39+2.55)*43.3 to get length
When you calculated the time it took to fall vertically, why is s negative? Also, why is a negative? Isn't a positive when an object is falling downwards? Thanks
It all depends on how you decide it in a question. I usually choose that everything upwards is positive, and everything downwards is negative. If you did it your way (so long as you're consistent always), you should be just fine :)
Ahh, that's because in the VERTICAL direction, we don't know v (the final vertical speed), so we can't use v=u+at to solve for t. You have to use vertical components only.
I don't know which smax equation you are referring to. I always recommend you just use the equations on the IB's data booklet so you are used to what you are given.
I'm using a Wacom Intuos tablet to draw, I'm using an older version of Baiboard as my whiteboard, and I'm using Camtasia to record my screen. Hope that helps!
Because it went DOWN 100m overall, so displacement is -100. Had it gone up 100 m, I would have made it positive. This trick is really helpful when solving : up and right are positive, and down and left I make negative. I like using this because it's the same thing we do in maths with graphs.
Hi! Good question - in physics and mathematics, we often put a small arrow to denote that a quantity is a vector instead of a scalar. A vector is something where we care about not only it's value, but also it's direction. A scalar we just care about it's value. Example: time is just a scalar, like t = 5 sec (it makes no sense to say t = 5 sec North!??) Example: velocity is a vector, like v = 5 m/s North (we care where it's pointing)
We often use the convention that in 2D projectiles, if something is going UP, then we use positive values (speed, displacement, etc). And if something is acting DOWNwards, then we use a negative. And, since the acceleration due to gravity always acts downwards, then we use -9.81 ms^-2
Ahh, you shouldn't use this formula for two reasons: 1. That formula is only valid for situations where the launch and landing are at the same height (both launch and impact on a horizontal surface). However, in this case, we're launching off the TOP of a cliff, but it will land much lower than the initial launch height (it lands on the ground below) 2. That formula isn't one I recommend for IB physics. Stick to the formulae they give you (assuming you're an IB diploma student). If you use the 4 equations of accelerated motion carefully every time, you can solve 2D projectile motion questions well.
Initial velocity in the y-direction is 25, and that's what you needed. That's why the first thing to do is break up the vector into components to find what you need.
I use negatives for directions that are down or to the left. This helps to make all the values make sense to me, plus it works with the equations. Hurray
There are a number of ways of solving that final quadratic. You could use the quadratic equation. Or, you could graph the left side of the equation and use your calculator to find the zeros. Or, you can use a polynomial root finder. Ex: the TI-84 has an app called Polysmlt 2. Use that to get the roots. Or you can use a similar function on your calculator (TI-nSpire can do it easily)
@@OSC1990 I used the polynomial root finder of TI-nspire and I got two roots (-2.6 and 7.7) I know since time can not be negative the answer is 7.7, but how about if we get 2 roots that are positive? Which one should we pick, or is that even possible to get 2 positive roots in these questions?
for part B, the final answer would be adding the displacement of x and the displacement of y right? Since its asking for the distance from the base of the cliff.
In part b, I'm ignoring what I just found in a), and starting again from the beginning with the initial conditions. Then you start with just a height of 100 m, so your displacement when you land in will be -100 m. Keep in mind, this is only in the VERTICAL direction we're considering.
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![Accelerated motion – graphing [IB Physics SL/HL]](/img/tr.png)
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"if you forgot to turn your calculator into degrees mode, you're gonna have a RAD time" lmao
yea how funny you damn nerd
@@iyadalahed9150 Dude his profile name tells he is nerd
in last question, seperate into two parts ( ascend and fall), 2.55 sec for ascend, v=0+2(-9.18)(-132), s=vt/2, then u get t for fall: 5.39. then (5.39+2.55)*43.3 to get length
that's what I do as well. Makes me feel like I'm being more thorough and thus reducing the chances of a mistake.
you saved me i was feeling completely lost in how to solve problems now I feel reassured
That's wonderful - I'm so glad you can benefit from the video! Cheers, Mitch
he's not even gonna teach us how to survive the impact
Hah!
When you calculated the time it took to fall vertically, why is s negative? Also, why is a negative? Isn't a positive when an object is falling downwards? Thanks
It all depends on how you decide it in a question. I usually choose that everything upwards is positive, and everything downwards is negative. If you did it your way (so long as you're consistent always), you should be just fine :)
Nice video. Thanks for the tutorial!
You are a saviour
for part b, why can't i use the equation v=u+at???? i get t=2.6 seconds...
Ahh, that's because in the VERTICAL direction, we don't know v (the final vertical speed), so we can't use v=u+at to solve for t. You have to use vertical components only.
why does the smax equation does not work to prove the last question?
I don't know which smax equation you are referring to. I always recommend you just use the equations on the IB's data booklet so you are used to what you are given.
@@OSC1990 thank you, my teacher confused me
What is the app you're using in the video?
I'm using a Wacom Intuos tablet to draw, I'm using an older version of Baiboard as my whiteboard, and I'm using Camtasia to record my screen. Hope that helps!
thank you!!!! such a life saver!!!
Glad it helped! Cheers, Mitch
i do not understand why we used -100 for displacement
Because it went DOWN 100m overall, so displacement is -100. Had it gone up 100 m, I would have made it positive. This trick is really helpful when solving : up and right are positive, and down and left I make negative. I like using this because it's the same thing we do in maths with graphs.
@@OSC1990 thanks
Why does he put an arrow on top of velocity and displacement what does it mean?
Hi! Good question - in physics and mathematics, we often put a small arrow to denote that a quantity is a vector instead of a scalar. A vector is something where we care about not only it's value, but also it's direction. A scalar we just care about it's value.
Example: time is just a scalar, like t = 5 sec (it makes no sense to say t = 5 sec North!??)
Example: velocity is a vector, like v = 5 m/s North (we care where it's pointing)
in the second equation why is acceleration -9.81 while part of the is -9.81 and the other part is +9.81
We often use the convention that in 2D projectiles, if something is going UP, then we use positive values (speed, displacement, etc). And if something is acting DOWNwards, then we use a negative. And, since the acceleration due to gravity always acts downwards, then we use -9.81 ms^-2
Why can't we use the range formula u^2 x sin (2 x angle) / g here. I'm getting a different value for this formula but don't understand why it's wrong
Ahh, you shouldn't use this formula for two reasons:
1. That formula is only valid for situations where the launch and landing are at the same height (both launch and impact on a horizontal surface). However, in this case, we're launching off the TOP of a cliff, but it will land much lower than the initial launch height (it lands on the ground below)
2. That formula isn't one I recommend for IB physics. Stick to the formulae they give you (assuming you're an IB diploma student). If you use the 4 equations of accelerated motion carefully every time, you can solve 2D projectile motion questions well.
why in the first question the inicial velocity is 25 and not 50?
Initial velocity in the y-direction is 25, and that's what you needed. That's why the first thing to do is break up the vector into components to find what you need.
Hi, I don't get why Vy in question b is not zero since when u land there is no vertical motion
Because it’s the final velocity before reaching the ground. At the ground it becomes 0, before = speed due to gravity * time
thank you sir, really
But how did acceleration become a negative
I use negatives for directions that are down or to the left. This helps to make all the values make sense to me, plus it works with the equations. Hurray
You are a saviour.
Thank you so much 🙏 very helpful video 😊!
thank you so much
what did he put in his calculator? to get 7.7
There are a number of ways of solving that final quadratic. You could use the quadratic equation. Or, you could graph the left side of the equation and use your calculator to find the zeros. Or, you can use a polynomial root finder. Ex: the TI-84 has an app called Polysmlt 2. Use that to get the roots. Or you can use a similar function on your calculator (TI-nSpire can do it easily)
@@OSC1990 I used the polynomial root finder of TI-nspire and I got two roots (-2.6 and 7.7) I know since time can not be negative the answer is 7.7, but how about if we get 2 roots that are positive? Which one should we pick, or is that even possible to get 2 positive roots in these questions?
why does the final velocity in the second problem is not 0. the projectile hits the landing surface and it stops. why am i wrong?
think of it as the final velocity before it hits the surface and stops. so the velocity before 0.
Time in the second one should be 4.7, not 7.7...
i got 7.7 too
If it helps, when you graph the equation one of the zeros is 7.7.
for part B, the final answer would be adding the displacement of x and the displacement of y right? Since its asking for the distance from the base of the cliff.
in b) why do you use 100 for the height instead of 132?
In part b, I'm ignoring what I just found in a), and starting again from the beginning with the initial conditions. Then you start with just a height of 100 m, so your displacement when you land in will be -100 m. Keep in mind, this is only in the VERTICAL direction we're considering.
ohhhhh for max height I needed to add