A ball is thrown straight upward and rises to a maximum height of 20 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?

Answer 1

v = Vi - g t
at max height, v = 0
0 = Vi -gt
so t at top = Vi/g

How high does it go?
20 = Vi t - (1/2)g t^2
20 = Vi (Vi/g) - (1/2) gVi^2/g^2

20 = (1/2) Vi^2/g
40 (9.81) = Vi^2
Vi = 19.8 m/s

when does v = 19.8/2 or 9.9 m/s
v = Vi - g t
9.9 = 19.8 - 9.81 t
9.81 t = 9.9
t = 1 second