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+246x+100
y=−16x
2
+246x+100
Answers
Step-by-step explanation:
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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+246x+100
To Find : maximum height reached by the rocket
Solution:
y=-16x²+246x+100
dy/dx = -32x + 246
=> x = 246/32
d²y/dx² = - 32 < 0
Hence x = 246/32 = 123/16 will give max height
-16x²+246x+100
= -16(123/16)²+246(123/16)+100
= -15129/16 + 30258/16 + 1600/16
= (16729 / 16)
= 1,045.6 feet
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