A ball is thrown straight upward and rises to a maximum height of 23 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

m g h = (1/2)m Vi^2
9.81 (23) = (1/2) Vi^2
so
Vi = 21.2 m/s initial speed up

v = Vi - g t
10.6 = 21.2 - 9.81 t
t = 1.08 seconds

h = 0 + Vi t - (1/2)g t^2
= 21.2(1.08) - 4.9 (1.08)^2
= 17.2 meters

Answer 2

V^2 = Vo^2 + 2g*h
Vo^2 = V^2 - 2g*h.
Vo^2 = 0 + 19.6*23 = 450.8
Vo = 21.2 m.

h = (0.5Vo)^2-Vo^2)/2g
h = ((10.62)^2-(21.2^2))/-19.6=17.2 m.