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?

Question

Mathematics

Roderick Mullen

4 July, 06:08

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?

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Answers (1)

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Mckenzie Mclean · 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