# Kinematics - ball bouncing off vertical walls [closed]

So here's my best attempt to state the question:

A ball is thrown from a 100 meter tall tower, which has a vertical wall in front of it. The initial velocity of the ball is 10m/s and the distance between the tower and the wall is 10m. Find the distance from the foot of the tower that the ball falls. (The ball obviously comes down bouncing, alternating between tower and wall. g = 10m/s^2)

What I did:

sqr() means square root

Range of the ball I found to be 20sqr(5) Thus the ball does 4 bounces, and falls at a distance 20sqr(5) - 40 meters from the base of the tower. Any problem in this?

Also, this is a very.... ummm..... temporary sort of way to find it (if it is correct). Any other, more beautiful and elaborate way to find it?

Thanks :D

• in which direction is ball thrown – DSinghvi Jun 13 '14 at 17:41
• Well towards the wall. Doesn't really matter I suppose? You can take the tower to be on either side. – Gummy bears Jun 13 '14 at 17:45
• will it not be square root of (95^2 + 10^2) , because it will take 1 sec to reach wall by the time it will go down by 5 so height from foot of tower is equal to 95 – DSinghvi Jun 13 '14 at 17:49
• Isn't that too big an answer? The two are only 10 meters apart? – Gummy bears Jun 13 '14 at 17:59
• on 100m tall tower – DSinghvi Jun 13 '14 at 18:05

Assuming frictionless wall component of velocity in downward direction will not change on collision and velocity horizontal component will change its direction and magnitude remain same by momentum conservation. SO to come down it will take 100=1/2*g*t*t

implies time = sqrt(20)

[sqrt(20)]=4 where [x] denotes greatest integer lesser than or equal to x

it will have four collision as each collision time difference is 1s because 10m/10m/s therefore distance = 10(sqrt(20)-4)