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Ch 02: Motion Along a Straight Line
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 2, Problem 43b

A 75007500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.252.25 m/s2 and feels no appreciable air resistance. When it has reached a height of 525525 m, its engines suddenly fail; the only force acting on it is now gravity. How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes?

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First, calculate the velocity of the rocket at the moment the engines fail using the kinematic equation: v=u+at, where u is the initial velocity (0 m/s, since it starts from rest), a is the acceleration (2.25 m/s²), and t is the time taken to reach 525 m.
Use the kinematic equation s=ut+12at^2 to find the time t it takes to reach 525 m. Here, s is 525 m, u is 0 m/s, and a is 2.25 m/s².
Once the engines fail, the rocket will decelerate due to gravity. Use the equation v=u-gt to find the time it takes for the rocket to reach its peak height, where g is the acceleration due to gravity (9.81 m/s²).
Calculate the total time from engine failure to when the rocket crashes back to the launch pad. This involves finding the time to reach the peak height and then using the equation s=ut+12gt^2 to find the time it takes to fall back down from the peak height to the launch pad.
Finally, use the equation v=u+gt to calculate the velocity of the rocket just before it crashes into the launch pad, where u is the velocity at the peak height (0 m/s) and g is 9.81 m/s².

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). This concept is crucial for understanding the rocket's initial motion, as the upward acceleration is due to the force exerted by the engines, allowing us to calculate its velocity and position over time.
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Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration, allowing us to calculate displacement, velocity, and time. These equations are essential for determining the rocket's velocity at the moment the engines fail and for predicting its subsequent motion under the influence of gravity alone.
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Gravitational Force

Gravitational force is the attractive force exerted by the Earth on objects, causing them to accelerate downward at 9.81 m/s². After the rocket's engines fail, gravity becomes the sole force acting on it, influencing its descent back to the launch pad. Understanding this force is key to calculating the time and velocity of the rocket's fall.
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Related Practice
Textbook Question

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on the way down and passes a point 30.030.0 m below its starting point 5.005.00 s after it leaves the thrower's hand. Ignore air resistance. What is the initial speed of the egg?

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Textbook Question

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.901.90 s. You may ignore air resistance, so the brick is in free fall. How tall, in meters, is the building?

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Textbook Question

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.005.00 m/s, releases a sandbag at an instant when the balloon is 40.040.0 m above the ground (Fig. E2.442.44). After the sandbag is released, it is in free fall. What is the greatest height above the ground that the sandbag reaches?

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Textbook Question

A 75007500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.252.25 m/s2 and feels no appreciable air resistance. When it has reached a height of 525525 m, its engines suddenly fail; the only force acting on it is now gravity. What is the maximum height this rocket will reach above the launch pad?

1832
views
Textbook Question

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.005.00 m/s, releases a sandbag at an instant when the balloon is 40.040.0 m above the ground (Fig. E2.442.44). After the sandbag is released, it is in free fall. Compute the position and velocity of the sandbag at 0.2500.250 s and 1.001.00 s after its release.

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Textbook Question

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.901.90 s. You may ignore air resistance, so the brick is in free fall. What is the magnitude of the brick's velocity just before it reaches the ground?

2520
views