A rocket is launched from a tower the height of the rocket why in feet is related to the time after launch X in seconds by the given equation use this equation find the maximum height reached by the rocket to the nearest 10th of a foot y=-16x^2+152x+83

Step-by-step explanation:

Given the height of the rocket modelled by the equation;

y=-16x^2+152x+83

The rocket reaches its maximum height when the velocity of the rocket is zero.

Since velocity is the change in displacement with respect to time;

velocity v = dy/dx

dy/dx = -32x + 152

at dy/dx = 0

-32x + 152 = 0

-32x = -152

x = -152/-32

x = 4.75s

To get maximum height reached by the ball, we will substitute x = 4.75 into the modeled function as shown;

y(4.75) = -16(4.75)^2+152(4.75)+83

y(4.75) = -361+722+83

y(4.75) = 444m

Hence the maximum height reached by the ball is 444m

there is no graph so i cant sorry

step-by-step explanation: