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+193+128

Answer 1

x = 6.03 seconds

Explanation:

The equation of motion of the rocket is

y = -16x^2 + 193x + 128

We can determine its maximum height ymax when its vertical velocity y' reaches zero. Taking the derivative of y, we get

y' = -32x + 193

The value nen of x that makes y' will give us the time needed by the rocket to reach its maximum height ymax. So setting y' = 0 we get

x = 193/32 = 6.03 second

Bonus:

Putting this value of x into y, we'll get

y(x = 0.59s) = -(16 ft/s^2)(6.03 s)^2 + (193 m/s)(6.03 s) + 128 ft

= 710 ft