A juggler throws balls into the air. He throws one when the previous one is at its highest point. How high the balls rise if he throws m balls per second?

Question

anonymous · Answer

If "m" balls are thrown per second, the time taken for a ball to reach its maximum height will be 1/m seconds. How to get this? See that the next ball is thrown only when the previous ball reaches its maximum height. If 'm' balls were thrown in 1 second this means that each ball was attaining its maximum ht in 1/m seconds.

This was the main part. Now we can proceed to find maximum height in 2 ways-

a)
  We know for upward journey , t=1/m   a=-g
  v=u-gt
  final velocity ,v = 0 (at highest point)
  u=gt = g/m
  Now we can apply 
  h=ut-1/2 gt^2
  Putting the values of u,t, we will get  
  h= g/2m^2


b) 
   The second method uses a trick that time taken to reach the maximum ht is same as     time taken to fall down.
So, we will now consider the downward journey of ball which also takes 1/m seconds
We apply

h=ut+1/2gt^2
Here u=0 ,t=1/m
We will again get , h=g/2m^2