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=-16x^2+165x+78

Answer 1

Step-by-step explanation:

The height of the rocket y in feet is related to the time after launch, x in seconds, by the given equation i.e.

$y=-16x^2+165x+78$ ......(1)

It is required to find the maximum height reached by the rocket. For maximum height put $\dfrac{dy}{dx}=0$.

So,

$\dfrac{d(-16x^2+165x+78)}{dx}=0\\\\-32x+165=0\\\\32x=165\\\\x=5.15\ s$

Put x = 5.15 in equation (1).

$y=-16(5.15)^2+165(5.15)+78\\\\y=503.39\ m$

So, the maximum height reached by the rocket is 503.39 m.