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

Question

3 years ago

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thrown is given by the formula h(t)=56t-16t^2 . What is the maximum height of the ball?

1 answer:

Tju [1.3M] · Answer 1 · 2021-12-04T13:45:36+03:00

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