A tennis ball is released at 30 degrees. It reaches max height of 20 ft find initial speed

Answer 1

So we are to assume its initial velocity was 30 deg above the horizon.
verticalspeed=vsin30

The KE in the horizontal direction remains constant, but in the vertical
1/2 m v'^2+mgh=1/2 m (V sin30)^2
at the top, v' is zero, so
g*20=1/2 V^2 *1/4)

v^2=8g*20ft=160ft*32ft/s^2

v=sqrt (160*32) ft/sec=21.6ft/s

check that.

Answer 2

Y^2 = Yo^2 + 2g*h = 0.
Yo^2 - 64*20 = 0, Yo = 35.8 Ft/s.

Vo*sin30 = 35.8, Vo = 71.6 Ft/s.