Question

a ball is thrown vertically upward with an initial velocity of 80 feet per second. the distance s (in feet) of the ball from the ground after t seconds is s(t) =80t-16t^2. at what time t will the ball strike the ground? For what time t is the ball more than 96 feet above the ground

Answer 1

1) The ball strikes the ground when s(t) = 0. Solve for 80t-16t^2=0.
2) Solve for 80t-16t^2=96 or 80t-16t^2-96 = 0 to get your answer.

Answer 2

I hope that helped!

Answer 3

Thanks