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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 + 261x + 130

User Dejal
by
2.5k points

1 Answer

27 votes
27 votes

Answer

The highest height attained by the rocket = 1194.4 feet

Step-by-step explanation

The height of the rocket, y, in feet as a function of the time, x in seconds as

y = -16x² + 261x + 130

We are then asked to find the maximum height reached by the rocket

To do this, we would use the differentiation analysis to obtain the maximum of this function.

At maximum point for any function,

The first derivative = (dy/dx) = 0

The second derivative = (d²y/dx²) < 0

y = -16x² + 261x + 130

First derivative

(dy/dx) = -32x + 261 = 0

32x = 261

Divide both sides by 32

(32x/32) = (261/32)

x = 8.15625 s

We can then substitute this value of x (time) into the equation to get the maximum height (y)

y = -16x² + 261x + 130

At x = 8.15625 s,

y = -16 (8.15625)² + 261 (8.15625) + 130

= -1064.39 + 2128.78 + 130

= 1194.39

= 1194.4 feet to the nearest tenth.

Hope this Helps!!!

User Jakub Kriz
by
3.6k points
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