Question

A softball pitcher tosses a ball straight upward from a height of 3 feet above the ground with an initial velocity of 50 feet per second. The acceleration due to gravity is –16 ft/s².Which quadratic equation represents the height of the ball above the ground after t seconds?

h(t) = at2 + vt + h0
h(t) = 50t2 – 16t + 3
h(t) = –16t2 + 50t + 3
3 = –16t2 + 50t + h0
3 = 50t2 – 16t + h0

Answer 1

Use h(t) = h0 + v0 t + (1/2) a t^2. With g = 32 ft/s^2 downward, (1/2)a = -(1/2)(32) = -16, v0 = 50, h0 = 3, so

h(t) = -16t^2 + 50t + 3.