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Chapter 2: Problem 79

A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.50 \(\mathrm{m} / \mathrm{s}^{2} .\) At 20.0 \(\mathrm{s}\) after blastoff, the engines suddenly fail, and the rocket begins free fall. (a) How high above the launch pad will the rocket eventually go? (b) Find the rocket's velocity and acceleration at its highest point. (c) How long after it was launched will the rocket fall back to the launch pad, and how fast will it be moving when it does so?

Short Answer

Expert verified
(a) Maximum height: 627.55 m (b) Velocity at highest point: 0 m/s, acceleration: -9.81 m/s虏 (c) Total time: 30.2 s, speed upon return: 111.03 m/s

Step by step solution

01

Determine Initial Acceleration Phase Height

To find how high the rocket goes during the acceleration phase, use the formula for distance under constant acceleration: \( s = ut + \frac{1}{2}at^2 \). In this case, \( u = 0 \) (initial velocity), \( a = 2.50 \mathrm{m/s^2} \), and \( t = 20.0 \mathrm{s} \). Thus, the height during acceleration is: \( s = \frac{1}{2} \times 2.50 \times (20)^2 = 500\mathrm{m} \).
02

Calculate Velocity at Engine Failure Time

To find the velocity of the rocket when the engines fail, use the formula \( v = u + at \). With \( u = 0 \), \( a = 2.50 \mathrm{m/s^2} \), and \( t = 20.0 \mathrm{s} \), we find \( v = 0 + 2.50 \times 20 = 50 \mathrm{m/s} \).
03

Determine Additional Ascent Height During Free Fall

After the engines fail, the rocket continues upwards until its velocity reaches zero. Use the formula \( v^2 = u^2 + 2as \), with \( v = 0 \) (at the highest point), \( u = 50 \mathrm{m/s} \), \( a = -9.81 \mathrm{m/s^2} \) (acceleration due to gravity). Solving for \( s \), we get \( 0 = (50)^2 + 2(-9.81)s \), thus \( s = 127.55 \mathrm{m} \).
04

Calculate Total Maximum Height

Add the height from the constant acceleration phase to the height during free fall: \( 500 \mathrm{m} + 127.55 \mathrm{m} = 627.55 \mathrm{m} \). Hence, the maximum height is \( 627.55 \mathrm{m} \).
05

Find Velocity and Acceleration at the Highest Point

At the highest point, the velocity is \( 0 \mathrm{m/s} \) as the rocket momentarily stops. The acceleration is \( -9.81 \mathrm{m/s^2} \) because it is in free fall.
06

Determine Time to Fall to the Launch Pad

Use the formula \( v = u + at \) to find the time taken to reach the maximum height (zero velocity), with the upward velocity \( u = 50 \mathrm{m/s} \) and \( a = -9.81 \mathrm{m/s^2} \). Time \( t \) taken is \( t = \frac{-u}{a} = \frac{-50}{-9.81} \approx 5.1 \mathrm{s} \). The total time is \( 20 \mathrm{s} + 5.1 \mathrm{s} + 5.1 \mathrm{s} \approx 30.2 \mathrm{s} \) due to symmetry in ascent and descent.
07

Calculate Velocity at Launch Pad Return

Use the speed acquired after falling the full maximum height. With initial velocity \( 0 \mathrm{m/s} \) at the top, and falling through \( 627.55 \mathrm{m} \), use \( v^2 = u^2 + 2as \) with \( a = 9.81 \mathrm{m/s^2} \). Solve for \( v \): \( v^2 = 2(9.81)(627.55) \), yielding \( v \approx 111.03 \mathrm{m/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Acceleration
Constant acceleration means that a body experiences the same change in velocity over each equal time interval. For a rocket taking off, constant acceleration is crucial as it provides a straight upward thrust. This predictable motion allows us to calculate important parameters like velocity and distance using simple kinematic equations.

Here, the rocket's motion is governed by the equation for constant acceleration:
  • Distance traveled: \[s = ut + \frac{1}{2}at^2\]Where \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time elapsed. For the rocket, \(u = 0\) (since it starts from rest), \(a = 2.50\, \mathrm{m/s^2}\), and \(t = 20.0\, \mathrm{s}\).
  • The formula gives us the height achieved during this phase as \(500\, \mathrm{m}\).
Free Fall
Free fall begins the moment a rocket's engines cut off. At this point, the only force acting on the rocket is gravity, pulling it back to Earth at an acceleration of \(-9.81 \, \mathrm{m/s^2}\). It鈥檚 essential to understand that free fall doesn't mean the rocket stops moving upward immediately. It continues to ascend until its upward velocity decreases to zero.

During free fall, we continued to calculate how much higher the rocket would rise. This is done using the equation:
  • Velocity when engines stop: \(v = 50\, \mathrm{m/s}\)
  • To find the additional height, utilize: \[v^2 = u^2 + 2as\]With the final velocity \(v = 0\, \mathrm{m/s}\) (at the peak), and initial velocity \(u = 50\, \mathrm{m/s}\), providing us an additional height of \(127.55\, \mathrm{m}\).
Velocity Calculation
Calculating velocity at different points of a rocket's flight is vital for understanding its motion. The rocket undergoes two primary phases: powered ascent and free fall.

  • During powered ascent, we calculate the velocity at engine cut-off using: \[v = u + at\]With \(u = 0\, \mathrm{m/s}\), \(a = 2.50\, \mathrm{m/s^2}\), and \(t = 20.0\, \mathrm{s}\), results in \(v = 50\, \mathrm{m/s}\).
  • When the rocket reaches its maximum height, the velocity is \(0\, \mathrm{m/s}\) because it stops momentarily before descending.
  • As it returns to the launch pad, we calculate its velocity due to free fall using: \[v^2 = u^2 + 2as\]Starting from rest at the peak (\(u = 0\)), it falls a distance of \(627.55\, \mathrm{m}\) under gravity, achieving \(v \approx 111.03\, \mathrm{m/s}\) upon impact.
Distance Under Constant Acceleration
Understanding how to calculate distance while a body is under constant acceleration is key, especially in problems involving rocketry. This involves using the kinematic equations that relate initial velocity, time, acceleration, and distance.

For the rocket's ascent with engine power, we used:
  • \[s = ut + \frac{1}{2}at^2\]Knowing \(u = 0\, \mathrm{m/s}\), \(a = 2.50\, \mathrm{m/s^2}\), and \(t = 20.0\, \mathrm{s}\), the primary distance just before the engines stop is calculated as \(500\, \mathrm{m}\).
  • Adding free fall motion where the engine fails, the rocket continues to rise further until gravity stops it completely at \(127.55\, \mathrm{m}\), leading to a total height above the launch pad calculated as \(627.55\, \mathrm{m}\).

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Most popular questions from this chapter

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