Question

A catapult launches a boulder with an upward velocity of 120 ft/s. The height of the boulder, h,

Answer 1

Answer:

Reaches a maximum height of 235.00 feet in 3.75 seconds.

Step-by-step explanation:

The height of the boulder, h,  in feet after t seconds is given by the function is given by :

$h=-16t^2+120t+10$ .....(1)

For maximum height, put $\dfrac{dh}{dt}=0$

i.e.

$\dfrac{d(-16t^2+120t+10)}{dt}=0\\\\-32t+120=0\\\\32t=120\\\\t=3.75\ s$

Put t = 3.75 in equation (1). So,

$h=-16(3.75)^2+120(3.75)+10\\\\h=235\ \text{feet}$

So, the boulder's maximum height is 235 feets and it takes 3.75 s to reach to its maximum height.