Question

The engine of a rocket initially at rest on the ground is ignited, causing the rocket to rise vertically with a constant upward acceleration of magnitude a2g. At alitude L, the rocket's engine shuts off?

A) Derive an expression for the speed of the rocket at the moment the engine is shut off.
B) To what maximum height H does the rocket rise above its initial position?
C) Derive an expression for the total time the rocket is in the air, .e. from start until it hits the ground.
D) Sketch, qualitatively, position, velocity, and acceleration as functions of time. Identify the point where the engine shuts off and the highest point.

Answer 1

The questions relate to the kinematics of a rocket with upward acceleration and subsequent projectile motion. The solution involves calculating velocity at engine shut-off, the maximum altitude reached, total time in the air, and graphically representing the position, velocity, and acceleration over time.

Step-by-step explanation:

The student is asking questions that pertain to the kinematics and dynamics of a rocket subject to constant acceleration, and then to projectile motion after engine shut-off.

1. Speed of the rocket at engine shut-off: Using kinematic equations with upward acceleration a as 2g (where g is the acceleration due to gravity), and initial velocity u as 0 because the rocket starts from rest, the final velocity v at altitude L is given by the equation v = u + at, which simplifies to v = 2gL as u = 0 and time t can be obtained from L = (1/2)at².
2. Maximum height H above initial position: After engine shut-off, the rocket is a projectile. The maximum height can be found using the equation H = L + (v² / (2g)), where v is the final velocity at engine shut-off and L is the altitude where the engine shut off.
3. Total time in the air: The total time is the sum of the time taken to reach altitude L, and the time taken to rise from L to H and then fall back to the ground. This is found by solving the kinematic equations for the upward journey and the subsequent free fall.
4. Acceleration, velocity, and position: The acceleration is constant 2g until the engine shuts off, after which it becomes -g during the projectile motion phase. The velocity increases linearly until engine shut-off, then decreases until reaching zero at the peak, and increases in the downward direction during the fall. The position increases quadratically until the engine shut-off, and from there follows the trajectory of a projectile.

Answer 2

At the point where the rocket shuts off its speed is v = g×t =9.81×t

Step-by-step explanation:

The solution to this can be found by considering the equations of an object in free fall thus

S = ut + 0.5×a×t2

v = u + at

v2 = u2 - 2×a×S

Thus, with the acceleration a = 2g we have

Where v = final velocity,

u = initial velocity = 0m/s

t = time

a = acceleration = 2g

S = height = L

thus we have v = 0 + 2×g×t = 2gt

S = u×t + 0.5×a×t2

L = 0 + 0.5×2×g×t2 = gt2

Thus, at the time the Rocket’s Engine Shuts Off we have

The initial velocity u = 2gt

Thus the velocity v = u + at (where a = -g is the acceleration due to gravity) v = u – gt

= 2gt- gt = gt

At the point where the rocket shuts off its speed is = g×t =9.81×t

Where g = 9.81m/s2 from which the velocity is approximately 9.81 m/s in the first second