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 th e given equation. Using the equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second y=-16x^2+235x+133
1 answer:
Answer:
Step-by-step explanation:
Given tq equation of the height reached by the rocket as y=-16x^2+235x+133
The velocity of the object at its maximum height is zero
v = dy/dx = 0(at max height)
v =-32x +235
0 =-32x+235
x = 235/32
x = 7.34
Hence the rocket will reach its maximum height after 7.34seconds
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