Question

A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 36t + 10. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?

Answer 1

$h(t)=-16t^{2}+36t+10$
$h'(t)=-32t+36$
When h'(t) = 0 the maximum height is reached (turning point).
32t = 36
t = 1.13s
The maximum height is found by getting the value of h(1.13) as follows:$h(1.13)=-16(1.13)^{2}+(36\times1.13)+10=30.25 feet.$