Questions: A rocket is launched vertically from rest on a launch pad with a constant acceleration of 2.5 m / s^2. After 20 seconds the rocket motor is shut off. Determine: a) The height of the rocket when the engine is shut off. b) The velocity of the rocket when the engine shuts off. c) The velocity and acceleration of the rocket when it reaches its maximum height. d) The maximum height the rocket will attain. e) The total time the rocket is in the air when it hits the ground.

Solution Steps

The rocket is launched with a constant acceleration \( a = 2.5 \, \text{m/s}^2 \) for \( t = 20 \, \text{s} \).

The height \( h \) can be calculated using the equation: \[ h = \frac{1}{2} a t^2 \]

Substituting the given values: \[ h = \frac{1}{2} \times 2.5 \times (20)^2 = 500 \, \text{m} \]

The velocity \( v \) at the time the engine is shut off can be calculated using: \[ v = a t \]

Substituting the given values: \[ v = 2.5 \times 20 = 50 \, \text{m/s} \]

At maximum height, the velocity \( v = 0 \, \text{m/s} \).

The acceleration is due to gravity, \( a = -9.81 \, \text{m/s}^2 \).

Final Answer