Question

A ball is thrown from the top of a 50-ft building with an upward velocity of 24 ft/s. When will it reach its maximum height? How far above the ground will it be? Use the equation h(t)= -16t2 + 24t + 50
Please help me with this problem, thanks.

Answer 1

g = 32 ft/s^2 approximately
v = Vi - g t
at top v = 0
0 = 24 - 32 t
solve for t
at that top point
h = Hi + Vi t - 16 t^2
h = 50 + 24 t - 16 t^2 (oh sorry, you know that already)

Answer 2

Thank you!