214k views
0 votes
A rocket is launched from a tower. The height of the rocket, yin fest, isrelated to the time after launch, in seconds, by the given equation Usingthis equation, find the mastímum height reached by the rocket, to the nearesttenth of a foot.y= -16x^2 + 152x +83

1 Answer

6 votes

Answer

Maximum height reached by the rocket = 444 ft.

Step-by-step explanation

The height of the rocket, y, at any time x is given as

y = -16x² + 152x + 83

We are now asked to find the maximum height reached by the rocket.

For any function, the maximum value (height in this case) occurs at the point where the first derivative of the function is 0 and the second derivative of the function is negative.

That is,

(dy/dx) = 0

(d²y/dx²) < 0

y = -16x² + 152x + 83

(dy/dx) = -32x + 152

At maximum height, (dy/dx) = 0

-32x + 152 = 0

-32x = -152

Divide both sides by -32

(-32x/-32) = (-152/-32)

x = 4.75 seconds

So, this gives the time when the maximum height occurs.

To find this exact height, we will put this term for x in the original equation and solve for y when x = 4.75 s

y = -16x² + 152x + 83

y = -16 (4.75)² + 152 (4.75) + 83

y = -16 (22.5625) + 722 + 83

y = -361 + 722 + 83

y = 444.0 ft.

To confirm that this is indeed the maximum height reached by the rocket, we will compute the second derivative of the function.

y = -16x² + 152x + 83

(dy/dx) = -32x + 152

(d²y/dx²) = -32 < 0

This is indeed the maximum value of the given height function.

Hope this Helps!!!

User Idij
by
7.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.