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Chapter 3: Problem 51

A rocket is launched at an angle of \(53.0^{\circ}\) above the horizontal with an initial speed of \(100 \mathrm{~m} / \mathrm{s}\). The rocket moves for \(3.00 \mathrm{~s}\) along its initial line of motion with an acceleration of \(30.0 \mathrm{~m} / \mathrm{s}^{2}\). At this time, its engines fail and the rocket proceeds to move as a projectile. Find (a) the maximum altitude reached by the rocket, (b) its total time of flight, and \((c)\) its horizontal range.

Short Answer

Expert verified
(a) The maximum altitude reached by the rocket is the sum of the height obtained in Step 1 and step 3. (b) The total time of flight is the sum of the powered flight time and the time of descent calculated in Step 4. (c) The horizontal range is calculated in Step 6 by multiplying the total flight time with the horizontal component of the initial velocity.

Step by step solution

01

Calculating the final velocity and altitude after powered flight

First, the final velocity at the end of the powered flight is calculated using the equation of motion \(v_f = v_i + at\), where \(v_i\) is the initial velocity, \(a\) is the acceleration and \(t\) is the time. The initial velocity can be broken down into vertical and horizontal components using the angle provided. The vertical component of initial velocity \(v_{i_y} = v_i \sin(53掳)\). The horizontal component of initial velocity \(v_{i_x} = v_i \cos(53掳)\). Here, the acceleration is only vertical. Hence, the final vertical velocity can be given as \(v_{f_y} = v_{i_y} + at\). The altitude after powered flight is given by the equation \(h = v_{i_y}t + 0.5at^2\).
02

Calculating the time it takes to reach the maximum altitude after powered flight ends

In this phase, the rocket is under the influence of gravity alone. The time to reach the maximum height can be calculated using the equation of vertical velocity: \(v_{f_y} = v_{i_y} - gt\), where \(g\) is the acceleration due to gravity and is approximately \(9.8 m/s^2\). Solving this for time when the final velocity is 0 (at maximum height) gives \(t = v_{f_y}/g\). This is the time it takes to reach the maximum height after the powered flight ends.
03

Calculating the maximum altitude reached

The maximum height is the sum of the altitude obtained in step 1 and the additional height gained after the engines stopped. Since the upward velocity is slowing down at a steady rate due to gravity, the additional altitude can be calculated using the formula \(h = 0.5gt^2\).
04

Calculating the time of descent

To find the time it takes for the rocket to hit the ground, we can use the equation \(h = 0.5gt^2\), solved for time. referring to the maximum height calculated in step 3.
05

Calculating the total flight time

The total flight time is the sum of the time the rocket was powered and the time it takes to fall to the ground after reaching the maximum height.
06

Calculating the horizontal range

The horizontal range can be calculated by multiplying the total time in the air by the horizontal component of the initial velocity: \(Range = v_{i_x} \cdot Total~time\)

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Key Concepts

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

Rocket Trajectory
Rocket trajectories are fascinating because they combine elements from both physics and engineering. When a rocket is launched, it initially travels under the influence of its engines. This involves working against gravity and often includes a significant horizontal component.
The angle of launch plays a crucial role in determining the path the rocket will follow. In our example, the rocket is launched at an angle of 53 degrees. This means that the motion can be split into two components:
  • Vertical: Influences the height and duration of the flight.
  • Horizontal: Determines how far along the ground it will travel.
These components help in calculating the rocket's trajectory, essentially giving us a map of its path. As soon as the engines stop, the rocket moves as a regular projectile, purely under the influence of gravity. Therefore, understanding the initial powered part is key to predicting the overall trajectory.
Kinematics
Kinematics involves the study of motion without considering the forces that cause it. This is essential in breaking down the rocket鈥檚 journey into understandable parts.
For our rocket, we initially calculate using kinematics to break initial velocity into horizontal and vertical components using trigonometry:
  • \(v_{i_y} = v_i \cdot \sin(53^{\circ})\): vertical component impacting altitude.
  • \(v_{i_x} = v_i \cdot \cos(53^{\circ})\): horizontal component impacting range.
The rocket accelerates vertically due to its engines, and we use kinematic equations to find final velocities and other positions. These kinematic equations are:
  • \(v_f = v_i + at\), where \(a\) is acceleration.
  • \(h = v_i\cdot t + \frac{1}{2}at^2\) to find height.
This framework helps determine how fast the rocket moves and how high it goes in its powered phase.
Time of Flight
The time of flight is crucial in understanding how long the total journey lasts鈥攆rom launch to landing. It's derived from both phases of the rocket's motion: powered flight and projectile motion.
During the powered flight, the time is explicitly given (3 seconds) and used to determine initial conditions as the rocket transitions into projectile motion. For projectile motion, two main things must be considered:
  • Vertical ascent time after engines switch off, calculable using the formula \(t = \frac{v_{f_y}}{g}\), with \(g = 9.8m/s^2\) being the acceleration due to gravity.
  • Descent time from maximum altitude back to the ground, obtained by knowing the maximum height.
Total time of flight is the combination of powered flight duration and these ascent and descent times. This is key to then finding out how far the rocket travels horizontally, as it鈥檚 directly tied to the time it stays aloft.

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Most popular questions from this chapter

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strikes the floor at distance \(d\) from the base of the counter. If the height of the counter is \(h\), (a) find an expression for the time \(t\) it takes the cup to fall to the floor in terms of the variables \(h\) and \(g\). (b) With what speed does the mug leave the counter? Answer in terms of the variables \(d, g\), and \(h .(c)\) In the same terms, what is the speed of the cup immediately before it hits the floor? (d) In terms of \(h\) and \(d\), what is the direction of the cup's velocity immediately before it hits the floor?

One of the fastest recorded pitches in major-league baseball, thrown by Joel Zumaya in 2006, was clocked at \(101.0 \mathrm{mi} / \mathrm{h}\) (Fig. P3.22). If a pitch were thrown horizonLally with this velocity, how far would the ball fall vertically by the time it reached home plate, \(60.5 \mathrm{ft}\) away?

A man lost in a maze makes three consecutive displacements so that at the end of his travel he is right back where he started. The first displacement is \(8.00 \mathrm{~m}\) westward, and the second is \(13.0 \mathrm{~m}\) northward. Use the graphical method to find the magnitude and direction of the third displacement.

A hiker starts at his camp and moves the following distances while exploring his surroundings: \(75.0 \mathrm{~m}\) north, \(2.50 \times 10^{2} \mathrm{~m}\) east, \(125 \mathrm{~m}\) at an angle \(30.0^{\circ}\) north of east. and \(1.50 \times 10^{2} \mathrm{~m}\) south. (a) Find his resultant displacement from camp. (Take east as the positive \(x\) -direction and north as the positive \(y\) -direction.) (b) Would changes in the order in which the hiker makes the given displacements alter his final position? Explain:

A chinook salmon has a maximum underwater speed of \(3.58 \mathrm{~m} / \mathrm{s}\), but it can jump out of water with a spced of \(6.26 \mathrm{~m} / \mathrm{s}\). To move upstream past a waterfall, the salmon does not need to jump to the top of the fall. but only to a point in the fall where the water speed is less than \(3.58 \mathrm{~m} / \mathrm{s}\); it can then swim up the fall for the remaining distance. Because the salmon must make forward progress in the water, let's assume it can swim to the top if the water speed is \(3.00 \mathrm{~m} / \mathrm{s}\). If water has a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) as it passes over a ledge, how far below the ledge will the water be moving with a speed of \(9.00 \mathrm{~m} / \mathrm{s}^{2}\) (Note that water undergoes projectile motion once it leaves the ledge.) If the salmon is able to jump vertically upward from the base of the fall, what is the maximum height of waterfall that the salmon can clear?
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