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A level maths mechanics help

I need help on this question, and anything will be appreciated thank you :smile:

A ball A falls vertically from rest from the top of a tower 63m high. At the same time as A begins to fall, another ball B is projected vertically upwards from the bottom of the tower with speed 21m/s. The balls collide.
Find the distance of the point where the balls collide from the bottom of the tower.
For this we use suvat.

For Ball A:
s1 = ?
u = 0
v = x
a = 9.81
t = t

For Ball B:
s2 = ?
u = 21
v = x
a = -9.81
t = t

s = ut + 0.5at^2
s1 = 0 x t + 0.5 x 9.81 x t^2 = 4.905t^2
s2 = 21t + 0.5 x -9.81 x t^2 = -4.095t^2 + 21t

s1 + s2 = 63
4.905t^2 - 4.905t^2 + 21t = 63
21t = 63
t = 3 secs

Hope this helps! :smile:
Reply 2
Avatar for Ajmaster
Ajmaster
6
Well you need to find x (displacement), therefore you need to make two suvat equations. As both are launched at the same time, they meet at the same time after launch, you are going to use this to make one equation with one unknown.

So for A, it starts at rest: u = 0; it is falling so: a = -9.8 (We are taking the upward direction as positive); it meets B at time t: t = T and we are trying to find the distance from the bottom, therefore s = x-63

For B it starts at 21m/s: u = 21; It is going up so under gravity: a = -9.8; again we need s at time t: t = T and distance travelled upwards is x: s = x.

So for a as you have: u,a,t and s; you can use the equation s = ut + 1/2*at2 . Now for b it is the same: you have: u,a,t and s; so you can make the equation s = ut + 1/2*at2;

So for A equation is: x-63 = 0T + 1/2*(-9.8)T2
for B the equation is: x = 21T + 1/2*(-9.8)T2;

sub in x as B into A so that you can find the time T when this occurs (the post above states T=3) and then sub this value of T back into equation B so that you can find x, which should be 21*3+0.5*(-9.8)*9 = 63 - 44.1 = 18.9m = x
Original post by Ajmaster
Well you need to find x (displacement), therefore you need to make two suvat equations. As both are launched at the same time, they meet at the same time after launch, you are going to use this to make one equation with one unknown.

So for A, it starts at rest: u = 0; it is falling so: a = -9.8 (We are taking the upward direction as positive); it meets B at time t: t = T and we are trying to find the distance from the bottom, therefore s = x-63

For B it starts at 21m/s: u = 21; It is going up so under gravity: a = -9.8; again we need s at time t: t = T and distance travelled upwards is x: s = x.

So for a as you have: u,a,t and s; you can use the equation s = ut + 1/2*at2 . Now for b it is the same: you have: u,a,t and s; so you can make the equation s = ut + 1/2*at2;

So for A equation is: x-63 = 0T + 1/2*(-9.8)T2
for B the equation is: x = 21T + 1/2*(-9.8)T2;

sub in x as B into A so that you can find the time T when this occurs (the post above states T=3) and then sub this value of T back into equation B so that you can find x, which should be 21*3+0.5*(-9.8)*9 = 63 - 44.1 = 18.9m = x

Thank you so much, that’s so helpful! :smile:
Reply 4
Avatar for Jono*
Jono*
18
Haven't checked the other answers but usually you use 10ms-1 as acceleration due to gravity even tho it's actually closer to 9.81, perhaps it's different for other exam boards tho

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