Question

A boy throws n balls per second at regular time intervals. When the first ball reaches the maximum height he throws the second one vertically up. Find the maximum height reached by each ball.

Answer 1

Let initial velocity be u
final velocity v =0
time taken to reach the highest point t =1/n
v=u +at
0=u-g/n
-u=-g/n
u=g/n

s=ut-gt^2/2
s=u/n -g/2n^2
u=g/n
s=g/n^2 -g/2n^2
s=g/2n^2

thus maximum Height equals gravitational acceleration over two times number of balls per second squared.