A ball is thrown into the air with an upward velocity of 32 feet per second. Its height, h, in feet after t

seconds is given by the function h (t) = - 16t² + 32t + 6. What is the ball's maximum height? How long does it

take the ball to reach its maximum height? Round to the nearest hundredth, if necessary.

Reaches a maximum height of 22 feet after 1.00 second.

Reaches a maximum height of 22 feet after 2.00 seconds.

Reaches a maximum height of 44 feet after 2.17 seconds.

Reaches a maximum height of 11 feet after 2.17 seconds.

Question

Physics

Guest

14 January, 04:54

A ball is thrown into the air with an upward velocity of 32 feet per second. Its height, h, in feet after t

seconds is given by the function h (t) = - 16t² + 32t + 6. What is the ball's maximum height? How long does it

take the ball to reach its maximum height? Round to the nearest hundredth, if necessary.

Reaches a maximum height of 22 feet after 1.00 second.

Reaches a maximum height of 22 feet after 2.00 seconds.

Reaches a maximum height of 44 feet after 2.17 seconds.

Reaches a maximum height of 11 feet after 2.17 seconds.

+3

Answers (2)

Know the Answer?

Tania Sampson · Answer 1

Hello there.

A ball is thrown into the air with an upward velocity of 32 feet per second. Its height, h, in feet after t

seconds is given by the function h (t) = - 16t² + 32t + 6. What is the ball’s maximum height? How long does it

take the ball to reach its maximum height? Round to the nearest hundredth, if necessary.

Reaches a maximum height of 22 feet after 1.00 second.

Golden · Answer 2

Differentiate the expression, to obtain expression for velocity. Set velocity to 0, this when max height is reached. Obtain the tmax from that expression.

h (t) = - 16t² + 32t + 6

h' (t) = - 32t² + 32

0 = - 32t² + 32

t max = 1

hmax = - 16 (1) ² + 32 (1) + 6

hmax = 22

Therefore, first option is the correct answer.