Question

A ball is thrown straight up from the top of a hill 30 feet high with initial velocity of 72 ft/sec. How high above level ground will the ball get? (objects subjected to gravity adhere to s(t)=-16t^2+vot+so where s is the height of the object in feet, vo is the initial velocity and so is the initial height).

Answer 1

At max height: t = - V₀/ g
Plug V₀ = 72, S₀ = 30 and t to get the max height S(t)