Question

A person throws a ball straight up. He releases the ball at a height of 1.75 m above the ground and with a velocity of 12.0 m/s. Ignore the effects of air resistance. (a) How long until the ball reaches its highest point? (b) How high above the ground does the ball go?

Answer 1

When the ball reaches the highest point his velocity equals to zero:
v=v₀ - gt = 0
12 = 10t
t = 6/5 = 1.2 s
From the equation of the motion:
h =h₀ +v₀t - gt²/2 = 1.75 + 12 * 1.2 - 10* 1.2²/2 = 8.95 m.