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 = –16t^2 + 36t + 10. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball’s maximum height?

Answer 1

Step-by-step explanation:

We have,

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 :

$h=-16t^2 + 36t + 10$ ......(1)

Part (a) :

The maximum height reached by the ball is given by :

$\dfrac{dh}{dt}=0\\\\\dfrac{d(-16t^2 + 36t + 10)}{dt}=0\\\\-32t+36=0\\\\t=\dfrac{36}{32}\\\\t=1.125\ s$

Part (b) :

The maximum height of the ball is calculated by putting t = 1.125 in equation (1) such that :

$h=-16(1.125)^2 + 36(1.125)+ 10\\\\h=30.25\ m$