Ajuggler 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?

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

The answer is d) by requesting advice on science internet sites