A ball is thrown into the air with an upward velocity of 20 feet per second it's height h in feet after t seconds is given by the function h (t) = 16t^2+20t+2. How long does it take the ball to reach its maximum height? What is the ball's maximum height?

Question

Saul Massey

Mathematics

1 October, 23:02

A ball is thrown into the air with an upward velocity of 20 feet per second it's height h in feet after t seconds is given by the function h (t) = 16t^2+20t+2. How long does it take the ball to reach its maximum height? What is the ball's maximum height?

+1

Answers (1)

Know the Answer?

Wyatt Reilly · Answer 1

The extreme of the quadratic ax² + bx + c is found at x = - b / (2a). Your function has

a = 16

b = 20

so it will have an extreme value at t = - 20 / (2*16) = - 20/32 = - 5/8.

We assume you intend h (t) = - 16t² + 20t + 2 (with a leading minus sign), in which case the extreme value (maximum height) occurs at t = 5/8. The maximum height is

h (5/8) = (-16*5/8 + 20) (5/8) + 2 = 50/8 + 2 = 33/4 = 8.25

The ball reaches its maximum height after 5/8 = 0.625 seconds.

The ball's maximum height is 8 1/4 = 8.25 feet.