Question

The path of a model rocket can be represented by the equation h(t)=-t2+15t+16, where h(t) is the height, in feet, of the rocket at any given time, t. what is the height of the model rocket after 3 seconds from launch?

Answer 1

we are given

`h(t)=-t^2+15t+16`

where h(t) is height in feet

t is time in seconds

now, we can plug t=3 and find height

`h(3)=-(3)^2+15(3)+16`

`h(3)=-9+45+16`

`h(3)=52`

so,

the height of the model rocket after 3 seconds from launch is 52 feet.......Answer

Answer 2

The height is 52 feet.

Using t=3, we have:
-3² + 15(3) + 16 = -9 + 45 + 16 = 52