Question

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

Answer 1

Answer:

y = 841.25 feet

Step-by-step 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:

$y = -16x^2 + 212x + 139$ ....(1)

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

For maximum height, put dy/dx = 0

So,

$\dfrac{d(-16x^2 + 212x + 139)}{dx}=0\\\\-32x+212=0\\\\x=\dfrac{212}{32}\\\\x=6.625\ s$

Pu x = 6.625 in equation (1).

$y = -16(6.625)^2 + 212(6.625) + 139\\\\y=841.25\ feet$

So, the maximum height reached by the rocket is equal to 841.25 feet.