COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
3rd Edition
ISBN: 9781319453916
Author: Freedman
Publisher: MAC HIGHER
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Chapter 2, Problem 82QAP
To determine

(a)

The maximum altitude of the rocket.

Expert Solution
Check Mark

Answer to Problem 82QAP

The maximum altitude of the rocket is 1091.7m

Explanation of Solution

Given:

a1 = 2.00 m/s2

a2 = 3.00 m/s2

t1 = 15 sec

t2 = 12 sec

Formula used:

  H=ut+12at2

S = distance travelled

u = initial velocity

a = acceleration

t = time taken

Calculation:

During first stage, the height reached is,

  H1=ut1+12a1t12=0×4+12×2×162=256m

At the end of first stage speed is,

  v1=u+a1t1=0+2×16=32m/s

During second stage, the height reached is,

  H2=v1t2+12a2t22=32×12+12×3×122=600m

At the end of second stage speed is,

  v2=v1+a2t2=32+3×12=68m/s

At the end of second stage, the rocket will be under the effect of gravity, so it will continue to move upwards up to a point where its kinetic energy will be zero.

The time duration for reaching the maximum altitude from the end of second stage is,

  v3=v2gt30=68(9.8×t3)t3=689.8=6.938sec

The distance between the end of second stage and maximum altitude is,

  H3=v2t312gt32=68×6.93812×9.81×(6.938)2=235.7m

So, the maximum height is,

  H1=H1+H2+H3=256+600+235.7=1091.7m

To determine

(b)

The average speed and average velocity when the rocket fall back to launch pad.

Expert Solution
Check Mark

Answer to Problem 82QAP

The average speed is 44.69m/s and average velocity is 0.

Explanation of Solution

Given:

The maximum altitude is 1091.67m

Calculation:

Once rocket reaches the maximum height, it will start to fall and reach the launch pad on Earth surface.

The time duration taken by rocket to fall back is,

  H=v3t4+12×g×t421091.7m=0×t4+12×9.81×t42t4=14.91sec

Total time duration of flight,

  t=t1+t2+t3+t4=15+12+6.938+14.91=48.85sec

Total distance travelled is,

  Ht=H+H=2H=2×1091.7m=2183.4m

So average speed will be,

  S=2183.3448.85=44.69m/s

Since, the initial and final position of rocket is same, so the net displacement is zero. Therefore, the average velocity will be,

  v=Dit=0t=0m/s

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Chapter 2 Solutions

COLLEGE PHYSICS-ACHIEVE AC (1-TERM)

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