Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 2, Problem 49P

(a)

To determine

The time interval for which the rocket is in motion above the ground.

(a)

Expert Solution
Check Mark

Answer to Problem 49P

The time interval for which the rocket is in motion above the ground is 41.0 s.

Explanation of Solution

Initially the rocket moves vertically upward with initial speed 80.0 m/s and acceleration 4.00 m/s2 . Then the engine fails and the rocket starts to move upward with acceleration equal to the negative of acceleration due to gravity until it reaches the maximum altitude. After this the rocket comes back to ground with acceleration equal to the negative of acceleration due to gravity Take the ground level to be point 0, point 1 to be at the end of the engine burn, point 2 to be the highest altitude reached by the rocket and point 3 to be at just before the impact on the ground.

The diagram is shown below.

Principles of Physics: A Calculus-Based Text, Chapter 2, Problem 49P

Write the kinematics equation.

  vf2=vi2+2aΔy        (I)

Here, vf is the final speed, vi is the initial speed, a is the acceleration and Δy is the displacement.

Apply equation (I) for the motion of the rocket from point 0 to point 1.

  vf012=vi012+2a01Δy01

Here, vf01 is the speed rocket when the engine fail, vi01 is the initial speed of the rocket at the launch, a01 is the acceleration produced by engine thrust and Δy01 is the displacement from point 0 to point 1 .

Substitute 80.0 m/s for vi01, 4.00 m/s2 for a01 and 1000 m for Δy01 in the above equation to find vf01.

  vf012=(80.0 m/s)2+2(4.00 m/s2)(1000 m)=14400 m2/s2vf01=120 m/s

Write the kinematics equation.

  vf=vi+at        (II)

Here, t is the time interval.

Apply equation (II) for the motion of the rocket from point 0 to point 1 .

  vf01=vi01+a01t01

Here, t01 is the time taken by the rocket to reach point 1 from point 0 .

Rewrite the above equation for t01.

  t01=vf01vi01a01

Substitute 120 m/s for vf01, 80.0 m/s for vi01 and 4.00 m/s2 for a01 in the above equation to find t01.

  t01=120 m/s80.0 m/s4.00 m/s2=10.0 s

Apply equation (I) for the motion of the rocket from point 1 to point 2.

  vf122=vf012+2(g)Δy12=vf0122gΔy12

Here, vf12 is the speed rocket at the maximum altitude, g is the acceleration due to gravity and Δy12 is the displacement from point 1 to point 2.

Rewrite the above equation for Δy12 .

  Δy12=vf122vf0122g

At the maximum altitude the speed of the rocket will be zero.

Substitute 0 for vf12, 120 m/s for vf01 and 9.80 m/s2 for g in the above equation to find Δy12.

  Δy12=0(120 m/s)22(9.80 m/s2)=735 m

Apply equation (II) for the motion of the rocket from point 1 to point 2.

  vf12=vf01+(g)t12=vf01gt12

Here, t12 is the time taken by the rocket to reach point 2 from point 1.

Rewrite the above equation for t12 .

  t12=vf12vf01g

Substitute 0 for vf12, 120 m/s for vf01 and 9.80 m/s2 for g in the above equation to find t12.

  t12=0120 m/s9.80 m/s2=12.2 s

Apply equation (I) for the motion of the rocket from point 2 to point 3 .

  vf232=vf122+2(g)Δy23=vf1222gΔy23        (III)

Here, vf23 is the speed with which the rocket strikes the ground and Δy23 is the distance from point 2 to point 3.

Write the equation for Δy23.

  Δy23=(Δy01+Δy12)

Substitute 1000 m for Δy01 and 735 m for Δy12 in the above equation to find Δy23.

  Δy23=(1000 m+735 m)=1735 m

Substitute 0 for vf12, 9.80 m/s2 for g and 1735 m for Δy23 in equation (III) to find vf23.

  vf232=02(9.80 m/s2)(1735 m)=34006 m2/s2vf23=184 m/s

The velocity is negative since it is directed downward.

Apply equation (II) for the motion of the rocket from point 2 to point 3.

  vf23=vf12+(g)t23=vf12gt23

Here, t23 is the time taken by the rocket to reach point 3 from point 2.

Rewrite the above equation for t23 .

  t23=vf23vf12g

Substitute 184 m/s for vf23, 0 for vf12 and 9.80 m/s2 for g in the above equation to find t23.

  t23=184 m/s09.80 m/s2=18.8 s

Write the equation for the time interval the rocket is in motion above the ground.

  ttot=t01+t12+t23        (IV)

Here, ttot is the time interval the rocket is in motion above the ground.

Conclusion:

Substitute 10.0 s for t01, 12.2 s for t12 and 18.8 s for t23 in equation (IV) to find ttot .

  ttot=10.0 s+12.2 s+18.8 s=41.0 s

Therefore, the time interval for which the rocket is in motion above the ground is 41.0 s.

(b)

To determine

The maximum altitude of the rocket.

(b)

Expert Solution
Check Mark

Answer to Problem 49P

The maximum altitude of the rocket is 1.73 km.

Explanation of Solution

Write the equation for the maximum altitude.

  Δy23=Δy01+Δy12

Conclusion:

Substitute 1000 m for Δy01 and 735 m for Δy12 in the above equation to find Δy23.

  Δy23=1000 m+735 m=1735 m1 km1000 m=1.735 km1.73 km

Therefore, the maximum altitude of the rocket is 1.73 km.

(c)

To determine

The velocity of the rocket just before it hits the ground.

(c)

Expert Solution
Check Mark

Answer to Problem 49P

The velocity of the rocket just before it hits the ground is 184 m/s.

Explanation of Solution

Equation (III) gives the velocity with which the rocket strikes the ground.

Conclusion:

Substitute 0 for vf12, 9.80 m/s2 for g and 1735 m for Δy23 in equation (III) to find vf23.

  vf232=02(9.80 m/s2)(1735 m)=34006 m2/s2vf23=184 m/s

Therefore, the velocity of the rocket just before it hits the ground is 184 m/s.

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Chapter 2 Solutions

Principles of Physics: A Calculus-Based Text

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