Question

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 out the time at which the rocket will reach its max, to the nearest 100th of a second. y=-16x^2+218x+75

Answer 1

Answer:

6.81s

Step-by-step explanation:

First you must know that the velocity of the body is zero at the maximum height.

v =dy/dx

v= -32x+218

Since v = 0

-32x+218 =0

32x =218

x= 218/32

x = 6.81s

Hence the rocket reaches its maximum height after 6.81s