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+161x+108
y=−16x
2
+161x+108

Answer 1

5.03secs

Explanation:

Given the height reached by the rocket expressed as;

y = -16x^2 + 161x + 108

The rocket reaches it max height when its velocity is zero

Velocity is expressed as;

V(x) = dy/dx = 0

dy/dx = -32x + 161

0 = -32x + 161

32x = 161

x = 161/32

x = 5.03

Hence it will take 5.03secs for the rocket to reach its maximum height