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+152x+83
Answers
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. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second. y=-16x^2+143x+121
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Given : 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
y=-16x²+152x+83
To Find : maximum height reached by the rocket
Solution:
y=-16x²+152x+83
dy/dx = -32x + 152
=> x = 152/32
d²y/dx² = - 32 < 0
Hence x = 152/32 = 19/4 will give max height
-16x²+152x+83
= -16(19/4)²+152(19/4)+83
= -361 + 722 + 83
= 361 + 83
= 444 feet
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