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

Answer:

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

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