Question

Suppose you drop a ball onto the ground (starting it with zero initial velocity). The ball bounces back upward with half the speed at which it hit the ground. Determine how high the ball bounches back as a percentage of the initial height. Derive an equation using variables.

Answer 1

@Vocaloid (:

Answer 2

\[v^2 _{f} = v^2 _{i} + 2 a d\]

I know we are dealing with this kinematic as there is no mention of time.

Answer 3

hm. I'm not 100% sure on this solution but I have some ideas so far

Answer 4

vf^2 = vi^2 + 2ad (where d is the initial height)

initial velocity is 0 so

vf^2 = 2ad

vf = sqrt(2ad) is the velocity at which it hits the ground

Answer 5

alright I got it let me just take a picture

Answer 6

so the key is to write the upward velocity in terms of the upward height not the original height

vu = sqrt(2gh_u) using the same formula vf^2 = vi^2 + 2ad except applied to the upward velocity only.

after that you just want to cancel out all variables except hd and hu to get the upward height in terms of the downward height

Answer 7

they want it as a percentage not a fraction so 25% of the org height

Answer 8

wow that looks so clean.

Answer 9

Thank you Vocaloid xD