Question

A ball is thrown straight up with a initial velocity of 56 feet per second . The height , h , of the ball t seconds after it is thrown is given by the formula h(t)=56t-16t^2 . What is the maximum height of the ball?

Answer 1

Formula: H(t) = 56t – 16t^2  

H(t) = - 16t^2 + 56t

A.      What is the height of the ball after 1 second? H (1) = 56(1) – 16(1) ^2 = 40 pt.

B.      What is the maximum height? X = - (56)/2(- 16) = 1.75 sec h (1.75) = 56(1.75) – 16(1.75) ^2 h (1.75) = 49ft.

C.      After how many seconds will it return to the ground? – 16t^2 + 56t = 0   - 8t =0 t = 0

-          8t (2 + - 7) = 0     2t – 7 = 0  t = 7/2   Ans: 3.5 seconds