Question

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the

time after launch, x in seconds, by the given equation. Using this equation, find the
maximum height reached by the rocket, to the nearest tenth of a foot.
y = -16x2 + 136x + 57

Answer 1

Maximum height = 346 feet

Explanation:

Given that,

The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation as follows :

`y = -16x^2 + 136x + 57` .....(1)

We need to find the maximum height reached by the rocket. For maximum height,

Put dy/dt = 0

So,

`(d)/(dt)( -16x^2 + 136x + 57)=0\\\\-32x+136=0\\\\x=(136)/(32)\\\\x=4.25`

No, put x = 4.25 in equation (1).

`y = -16(4.25)^2 + 136(4.25)+ 57\\\\y=346\ feet`

So, the maximum height reached by the rocket is equal to 346 feet.