Question

Xavier shoots a basketball in which the height, in feet, is modeled by the equation, h(t) = -4t^2+10t +18, where t is time, in

seconds. What is the maximum height of the basketball?

Answer 1

h = 24.25 feet

Explanation:

Xavier shoots a basketball in which the height, in feet, is modeled by the equation.

`h(t) = -4t^2+10t +18` ....(1)

For maximum height,

Put dh/dt=0

So,

`(dh)/(dt)=(d)/(dt)(-4t^2+10t +18)\\\\=-8t+10`

So,

-8t = -10

t =1.25s

Put t = 1.25 s in equation (1)

`h(t) = -4(1.25)^2+10(1.25) +18\\\\=24.25\ feet`

So, the maximum height of the basketball is equal to 24.25 feet.