Question

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.

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.

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.