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Chapter 2: Problem 75

A pollution-sampling rocket is launched straight upward with rockets providing a constant acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2}\) for the first \(1000 \mathrm{~m}\) of flight. At that point the rocket motors cut off and the rocket itself is in free fall. Ignore air resistance. (a) What is the rocket's speed when the engines cut off? (b) What is the maximum altitude reached by this rocket? (c) What is the time it takes to get to its maximum altitude?

Short Answer

Expert verified
(a) 154.92 m/s, (b) 2223.62 m, (c) 28.70 s.

Step by step solution

01

Calculate Speed at Engine Cut-Off

Use the kinematic equation for velocity: \( v^2 = u^2 + 2as \), where \( v \) is the final speed, \( u \) is the initial speed (0 \( \mathrm{m/s} \) since the rocket starts from rest), \( a \) is the acceleration (\( 12.0 \mathrm{~m/s^2} \)), and \( s \) is the displacement (\( 1000 \mathrm{~m} \)). Plugging in the values, we get: \[ v^2 = 0 + 2 \times 12.0 \times 1000 \] \[ v^2 = 24000 \] \[ v = \sqrt{24000} = 154.92 \mathrm{~m/s} \]. Thus, the rocket's speed when the engines cut off is \( 154.92 \mathrm{~m/s} \).
02

Calculate Maximum Altitude

To find the maximum altitude, we need to consider the motion after the engines cut off. The initial speed is now \( 154.92 \mathrm{~m/s} \), and the rocket is subject only to gravity (\( -9.81 \mathrm{~m/s^2} \)). Using the kinematic equation \( v^2 = u^2 + 2as \) where final velocity \( v = 0 \) at the highest point, we can solve for additional height \( s \): \[ 0 = (154.92)^2 - 2 \times 9.81 \times s \] \[ s = \frac{(154.92)^2}{2 \times 9.81} = 1223.62 \mathrm{~m} \]. The total maximum altitude is \( 1000 \mathrm{~m} + 1223.62 \mathrm{~m} = 2223.62 \mathrm{~m} \).
03

Calculate Time to Reach Maximum Altitude

First, calculate the time during powered flight using the equation \( v = u + at \), where \( v = 154.92 \mathrm{~m/s} \), \( u = 0 \), and \( a = 12.0 \mathrm{~m/s^2} \): \[ 154.92 = 0 + 12.0t \] \[ t = \frac{154.92}{12.0} = 12.91 \mathrm{~s} \]. Next, calculate the time during free fall using \( v = u + at \), where \( v = 0 \), \( u = 154.92 \mathrm{~m/s} \), and \( a = -9.81 \mathrm{~m/s^2} \): \[ 0 = 154.92 - 9.81t \] \[ t = \frac{154.92}{9.81} = 15.79 \mathrm{~s} \]. Total time to reach maximum altitude is \( 12.91 \mathrm{~s} + 15.79 \mathrm{~s} = 28.70 \mathrm{~s} \).

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Key Concepts

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

Projectile Motion
Projectile motion occurs when an object is thrown or projected into the air, subject to the acceleration due to gravity. In the problem involving the pollution-sampling rocket, it's important to understand how the rocket behaves during launch and ascent. Initially, the rocket is propelled upwards by its engines, providing a constant acceleration. However, once the engines cut off, the rocket's motion is governed by projectile motion principles, primarily affected by gravity.

When analyzing projectile motion, you need to consider:
  • Initial velocity: The speed at which the rocket or object is projected upwards or at an angle.
  • Acceleration due to gravity: A constant downward force that affects the motion of the projectile. Typically, this is approximately \(9.81 \, \mathrm{m/s^2}\) on Earth.
  • Displacement: The overall change in position of the projectile.
  • Time of flight: The total time the rocket stays in the air, which includes both the powered and free fall phases in this scenario.
Understanding these elements aids in calculating the rocket's speed at different stages and its maximum altitude.
Constant Acceleration
Constant acceleration means that the velocity of an object changes by the same amount each second. In physics, a constant acceleration is typically due to gravity, but it can also result from other forces, like the rocket engines in the exercise.

For the first part of the flight, the rocket enjoys constant acceleration of \(12.0 \, \mathrm{m/s^2}\), thanks to its engines. During this phase, kinematic equations can be utilized to determine various aspects like:
  • Final velocity (\(v\)): Calculated using \(v = u + at\) or the equivalent velocity-squared equation: \(v^2 = u^2 + 2as\).
  • Displacement (\(s\)): The distance covered in the powered flight, which is \(1000 \, \mathrm{m}\) in this problem.
  • Time (\(t\)): The duration of time under the influence of a constant force before the engines cut off.
Such consistent acceleration simplifies the calculation of the rocket's speed and flight dynamics, crucial for solving physics problems like determining maximum altitude or total flight time.
Physics Problems
Physics problems often challenge you to apply theoretical concepts to real-world situations. In analyzing rocket flight physics, understanding kinematic equations becomes vital. These equations help solve problems involving objects in motion, especially under constant acceleration.

The problem can be approached step-by-step:
  • Initial Values: Begin with what you know - initial velocity, acceleration, and displacement.
  • Equation Application: Use relevant equations to find unknowns like velocity or time. Each phase of motion鈥攑owered flight and free fall鈥攎ight require separate calculations.
  • Problem Solving: Break complex calculations into manageable parts. For the rocket, first solve for velocity at engine cut-off, then follow with maximum altitude and total time calculations.
By applying these techniques, physics problems transform into solvable puzzles rather than intimidating mysteries. Grasping the sequential approach ensures you confidently navigate through any physics challenge, be it on exams or real-life scenarios.

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Most popular questions from this chapter

A regional airline flight consists of two legs with an intermediate stop. The airplane flies \(400 \mathrm{~km}\) due north from airport A to airport B. From there, it flies \(300 \mathrm{~km}\) due east to its final destination at airport \(C\). (a) What is the plane's displacement from its starting point? (b) If the first leg takes \(45 \mathrm{~min}\) and the second leg \(30 \mathrm{~min}\), what is the average velocity for the trip? (c) What is the average speed for the trip? (d) Why is the average speed not the same as the magnitude for the average velocity?

A photographer in a helicopter ascending vertically at a constant rate of \(12.5 \mathrm{~m} / \mathrm{s}\) accidentally drops a camera out the window when the helicopter is \(60.0 \mathrm{~m}\) above the ground. (a) How long will the camera take to reach the ground? (b) What will its speed be when it hits?

Many highways with steep downhill areas have "runaway truck" inclined paths just off the main roadbed. These paths are designed so that if a vehicle's braking system gives out, the driver can steer it onto this incline (usually composed of loose gravel or sand). The idea is that the vehicle can then roll up the incline and come permanently and safely to rest with no need of a braking system. In one region of Hawaii the incline distance is \(300 \mathrm{~m}\) and provides a (constant) deceleration of \(2.50 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the maximum speed that a runaway vehicle can have as it enters the incline? (b) How long would such a vehicle take to come to rest? (c) Suppose another vehicle moving \(10 \mathrm{mi} / \mathrm{h}(4.47 \mathrm{~m} / \mathrm{s})\) faster than the maximum value enters the incline. What speed will it have as it leaves the gravel-filled area?

Fig. \(2.25,\) a student at a window on the second floor of a dorm sees his math professor walking on the sidewalk beside the building. He drops a water balloon from \(18.0 \mathrm{~m}\) above the ground when the professor is \(1.00 \mathrm{~m}\) from the point directly beneath the window. If the professor is \(1.70 \mathrm{~m}\) tall and walks at a rate of \(0.450 \mathrm{~m} / \mathrm{s},\) does the balloon hit her? If not, how close does it come?

A train makes a round trip on a straight, level track. The first half of the trip is \(300 \mathrm{~km}\) and is traveled at a speed of \(75 \mathrm{~km} / \mathrm{h}\). After a \(0.50 \mathrm{~h}\) layover, the train returns the \(300 \mathrm{~km}\) at a speed of \(85 \mathrm{~km} / \mathrm{h}\). What is the train's (a) average speed and (b) average velocity?
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