Question

A ball is dropped from a building of height h. Assume the ball starts from rest and that air friction can be ignored. Derive an expression for the amount of time it takes the ball to travel from h to h/2.

Answer 1

Answer:

$t=\sqrt{h/g}$

Explanation:

We use the kinematics equation to solve this question:

$y(t)=y_{o}+v_{o}t+1/2*a*t^{2}$

$v_{o}=0$    because the ball is dropped

$a=-g$         the acceleration is the gravity, negative because it points downwards

$y_{o}=h$     initial height

$y(t)=h/2$     final height

So:

$h/2=h-1/2*g*t^{2}$

$t=\sqrt{h/g}$