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 + 254x + 79

Answer 1

The maximum height is 1087.06 ft.

Explanation:

We need to take the derivative of the equation and set it to 0.

`y' = - 32x + 254 = 0\\32x = 254\\x = 7.9375`

We now can plug x = 7.9375 into the original equation.

`y=-16x^(2) +254x+79\\y=-16(7.9375)^(2)+254(7.9375)+79\\y=1087.06 ft`