For tennis ball,
the boy throws it upwards.
s = ut + 1/2 at^2
a = -10 if I am taking upward as positive.
s = ut - 5t^2
If you have drawn the displacement vs time graph of the tennis
ball, it will be a 
curve, or
frowning curve. The maximum height of the path of tennis ball
implies the stationary point of the curve in this case.
Hence,
ds/dt = u - 10t
when ds/dt = 0,
u - 10t = 0
u = 10t where u represents the initial speed of tennis ball.
sub u =10t into s = ut - 5t^2
s = 10t^2 - 5t^2
= 5t^2
This means that at v = 0 which is the maximum height, displacement
travelled by the ball upwards = 5t^2.
For golf ball,
the boy drops it.
u = 0
a = 10
s = ut + 1/2 at^2
In this case, s is the displacement travelled by the golf ball
taking downwards to be positive.
s = 1/2(10)(t^2)
= 5t^2
This means that at time t seconds, the golf ball will travel 5t^2
downwards.
Since the balls clash at this point, assuming that the diameter of
the ball is negligible,
5t^2 + 5t^2 = 150
10t^2 = 150
t = root(15)
Initial speed of tennis ball = 10t
= 10root(15)
Position of tennis ball = 5t^2
= 5(15)
= 75 m
