College Physics:
College Physics:
11th Edition
ISBN: 9781305965515
Author: SERWAY, Raymond A.
Publisher: Brooks/Cole Pub Co
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Chapter 2, Problem 53P

A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s2 until its engines stop at an altitude of 150. m. (a) What can you say about, the motion of the rocket alter its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air?

(a)

Expert Solution
Check Mark
To determine
The motion of the rocket after the engines stop.

Answer to Problem 53P

After its engines stop, the rocket is a freely falling body under gravity. It continues upward and eventually slows under the influence of gravity. The rocket comes to rest momentarily at its maximum altitude. Then it falls back to Earth, gaining speed as it falls due to the acceleration due to gravity.

Explanation of Solution

The motion of the rocket can be determined using the acceleration of the rocket.

When the rocket moves upwards, then the engines forces the rocket to move upwards. When there is no external force acting on the rocket, the rocket is moving under the gravitational force. The magnitude of the acceleration acting on the rocket is the acceleration due to gravity. The acceleration due to gravity acts always downwards.

When the engine stops, the rocket starts moves under the acceleration due to gravity and which is opposite to the direction of the motion. This will slow down the rocket. As the rocket reaches its maximum height, the rocket has zero velocity. Then it falls under gravity. The acceleration and the motion of the rocket go on the same direction. This results in speeding up the rocket from zero to the maximum speed when it reaches the ground.

Conclusion:

After its engines stop, the rocket is a freely falling body under gravity. It continues upward and eventually slows under the influence of gravity. The rocket comes to rest momentarily at its maximum altitude. Then it falls back to Earth, gaining speed as it falls due to the acceleration due to gravity.

(b)

Expert Solution
Check Mark
To determine
The maximum height the rocket reaches.

Answer to Problem 53P

The maximum height the rocket reaches is 308m .

Explanation of Solution

Given info: The initial velocity of the of the rocket is 50.0m/s , the upward acceleration of the of the rocket is 2.00m/s2 , the height at which the engine stops is 150m , the velocity of the rocket at maximum height is 0 , and the acceleration of the rocket after the engine stops is 9.80m/s2 .

Explanation:

The formula used to calculate the final velocity of the rocket when it is accelerating upwards is,

v=v02+2aΔy

Here,

(Δy) is the height at which the engine stops

v0 is the initial velocity of the rocket

vf is the final velocity of the rocket at maximum height

a is the upward acceleration

Substitute 50.0m/s for v0 , 2.00m/s2 for a and 150m for Δy to find v .

v=(50.0m/s)2+2(2.00m/s2)(150m)=55.7m/s

Thus, The final velocity of the rocket when the engine stops is 55.7m/s .

The formula used to calculate the displacement of the rocket after the engine stops is,

Δy'=vf2v22a'

Here,

Δy' is the displacement of the rocket after the engine stops

vf is the final velocity of the rocket at maximum height

a' is the acceleration of the rocket after the engine stops

Substitute 0 for vf , 55.7m/s for v and 9.80m/s2 for a to find Δy' .

Δy'=(0)2(55.7m/s)22(9.80m/s2)=158m

Thus, The displacement of the rocket after the engine stops is 158m .

The formula used to calculate the maximum height the rocket reaches is,

hmax=Δy+Δy'

Here,

hmax is the maximum height the rocket reaches

Substitute 150m for Δy and 158m for Δy' to find hmax .

hmax=150m+158m=308m

Thus, the maximum height the rocket reaches is 308m .

Conclusion:

The maximum height the rocket reaches is 308m .

(c)

Expert Solution
Check Mark
To determine
The time taken by the rocket to reach the maximum height.

Answer to Problem 53P

The time taken by the rocket to reach the maximum height 8.51s .

Explanation of Solution

The formula used to calculate the interval at which the rocket moves with an upward acceleration is,

t1=Δy(v0+v2)

Here,

t1 is duration at which the rocket has upward acceleration

Substitute 150m for Δy , 50.0m/s for v0 and 55.7m/s for v to find t1 .

t1=150m((50.0m/s)+(55.7m/s)2)=2.84s

Thus, the time interval at which the rocket has upward acceleration is 2.84s .

The formula used to calculate the interval at which the rocket moves upwards after the engine stops is,

t2=Δy'(v+vf2)

Here,

t2 is duration at which the rocket moves upwards after the engine stops

Substitute 158m for Δy' , 55.7m/s for v and 0 for vf to find t2 .

t2=158m((55.7m/s)+(0)2)=5.67s

Thus, the time interval at which the rocket moves upward after the engine stops is 5.67s .

The formula used to calculate the total time of the upward flight is,

t=t1+t2

Here,

t is total time of upward flight

Substitute 2.84s for t1 and 5.67s for t2 to find t .

t=2.84s+5.67s=8.51s

Thus, the time taken by the rocket to reach the maximum height 8.51s .

Conclusion:

The time taken by the rocket to reach the maximum height 8.51s .

(d)

Expert Solution
Check Mark
To determine
The total time the rocket stays in air.

Answer to Problem 53P

The total time the rocket stays in air 16.4s .

Explanation of Solution

The formula used to calculate the total time of the downward flight is,

t3=2(hmax)a'

Here,

t3 is the total time of the downward flight is,

Substitute 308m for hmax and 9.80m/s2 for a' to find t3 .

t3=2(308m)(9.80m/s2)=7.93s

Thus, the total time of the downward flight is 7.93s .

The formula used to calculate the total time of flight is,

T=t+t3

Here,

T is the total time of flight of the rocket

Substitute 8.51s for t and 7.93s for t3 to find T .

T=8.51s+7.93s=16.4s

Thus, the total time the rocket stays in air 16.4s .

Conclusion:

The total time the rocket stays in air 16.4s .

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