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*1.13)+10=30.25 feet.`