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.

Answer 2

The maximum height of the rocket launched from a tower is 503.4 ft.

y represent the height of the rocket in feet and x represent the time after launch.

Given the relationship between y and x as:

y=-16x²+165x+78

To find the maximum height, the height is maximum at dy/dx = 0. Hence:

dy/dx = -32x + 165

-32x + 165 = 0

32x = 165

x = 5.156 seconds.

The maximum height is:

y= -16(5.156)² + 165(5.156) + 78

y = 503.4 ft.

The maximum height is 503.4 ft.

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