A small rocket is fired vertically upward, with an initial velocity of 60 m/s. A

Question

A small rocket is fired vertically upward, with an initial velocity of 60 m/s. Assuming, for simplicity, that gravitational acceleration is 10m/s^2, and neglecting all forces other than gravity, the rocket will experience a constant downward acceleration of 10 m/s^2. Give the rocket's upward velocity as a function of the time elapsed after the launch. Assuming that the rocket is launched from ground level, give its height above ground level as a function of the time elapsed after the launch. What is the maximum height reached by the rocket, and at what time will it be reached? At what time will the rocket hit the ground on the way down, and what will be its speed at impact?

Explanation / Answer

(a) u = 60, a = -10

v = u + at

v = 60 - 10t

(b) s = v dt = (60 - 10t) dt = 60t - 5t^2 + C

When t = 0, S = 0, so we get 0 = C

Therefore s = 60t - 5t^2

(c) When the maximum height is reached, v = 0

60 - 10t = 0, so t = 6 seconds

s = 60t - 5t^2 = 5t(12 - t) = 0 gives t = {0, 12}

The rocket will hit the ground 12 seconds after launch

v = u + at

v = 0 + 10(12) = 120 m/s

The impact velocity will be 120 m/s.

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