You throw a baseball directly upward at time t = 0 at an initial speed of 12.5 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9.80 m/s2.

Answer 1

V^2 = Vo^2 + 2g*h
h = (V^2-Vo^2)/2g
hmax = (0-12.5^2)/-19.6 = 8 m.

h = Vo*t + 0.5g*t^2 = 8/2 = 4 m
12.5*t + (-4.9)t^2 = 4
-4.9t^2 + 12.5t - 4 = 0

t = 0.375, and 2.18 s.
(Use Quadratic Formula).