Concept explainers
(a)
The speed of payload at the end of the stage 1.
Answer to Problem 78QAP
The speed of payload at the end of the stage 1 is 150 m/s
Explanation of Solution
Given:
Acceleration in stage 1,
Acceleration in stage 2,
Time to burn payload in stage 1,
Time to burn payload in stage 2,
Formula used:
u = initial velocity
t = time
a = acceleration
Calculation:
To find the velocity at the end of first stage, substitute the values in above equation,
(b)
The speed of payload at the end of the stage 2.
Answer to Problem 78QAP
The speed of payload at the end of the stage 2 is 210 m/s
Explanation of Solution
Given:
Acceleration in stage 1,
Acceleration in stage 2,
Time to burn payload in stage 1,
Time to burn payload in stage 2,
Formula used:
u = initial velocity
t = time
a = acceleration
Calculation:
To find the velocity at the end of second stage, substitute the values in above equation,
(c)
The altitude of the rocket at the end of stage 1.
Answer to Problem 78QAP
The altitude of the rocket at the end of stage 1 is 750 m.
Explanation of Solution
Given:
Acceleration in stage 1,
Acceleration in stage 2,
Time to burn payload in stage 1,
Time to burn payload in stage 2,
Formula used:
s = distance travelled.
u = initial velocity
t = time
a = acceleration
Calculation:
Substituting the values in above equation, we get-
(d)
The altitude of the rocket at the end of stage 2.
Answer to Problem 78QAP
The altitude of the rocket at the end of stage 2 is 1650 m
Explanation of Solution
Given:
Acceleration in stage 1,
Acceleration in stage 2,
Time to burn payload in stage 1,
Time to burn payload in stage 2,
Formula used:
s = distance travelled.
u = initial velocity
t = time
a = acceleration
Calculation:
Substituting the values in above equation, we get-
(e)
The maximum altitude obtained.
Answer to Problem 78QAP
The maximum altitude obtained is 3900 m.
Explanation of Solution
Given:
Acceleration in stage 1,
Acceleration in stage 2,
Time to burn payload in stage 1,
Time to burn payload in stage 2,
Formula used:
v = final velocity.
u = initial velocity.
g = acceleration due to gravity.
Calculation:
The extra height covered after stage 2 is,
So, the highest distance is,
(f)
The total travel time of the payload from launch to landing.
Answer to Problem 78QAP
The total travel time of the payload from launch to landing is 64.64 sec.
Explanation of Solution
Given:
Acceleration in stage 1,
Acceleration in stage 2,
Time to burn payload in stage 1,
Time to burn payload in stage 2,
Formula used:
s = distance travelled.
u = initial velocity
t = time
a = acceleration
Calculation:
Time taken to cover extra height is,
Time taken to drop back to ground is,
So, total time of flight is,
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Chapter 2 Solutions
COLLEGE PHYSICS,VOLUME 1
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