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Chapter 3: Problem 40

A test rocket is launched by accelerating it along a \(200.0-\mathrm{m}\) incline at \(1.25 \mathrm{~m} / \mathrm{s}^{2}\) starting from rest at point \(A\) (Figure 3.38 ). The incline rises at \(35.0^{\circ}\) above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the maximum height above the ground that the rocket reaches, and (b) the greatest horizontal range of the rocket beyond point \(A\).

Short Answer

Expert verified
Maximum height: 123.04 m; Horizontal range: 47.58 m.

Step by step solution

01

Calculate the Speed of Rocket at the End of Incline

First, we need to calculate the final speed of the rocket at the end of the 200 m incline. We'll use the kinematic equation for acceleration, which is given by: \[ v^2 = u^2 + 2a s \]where \(v\) is the final velocity, \(u\) is the initial velocity (0 m/s, since it starts from rest), \(a\) is the acceleration \((1.25 \text{ m/s}^2)\), and \(s\) is the distance \((200 \text{ m})\). Solving for \(v\), we have:\[ v^2 = 0 + 2 \times 1.25 \times 200 \]\[ v^2 = 500 \]\[ v = \sqrt{500} = 22.36 \text{ m/s} \]
02

Determine Components of Velocity

At the end of the incline, the rocket's velocity has both horizontal and vertical components due to the 35-degree angle. These components can be determined using trigonometry:\[ v_x = v \cdot \cos(35^\circ) \]\[ v_y = v \cdot \sin(35^\circ) \]Substituting \(v = 22.36 \text{ m/s}\),\[ v_x = 22.36 \cdot \cos(35^\circ) = 18.30 \text{ m/s} \]\[ v_y = 22.36 \cdot \sin(35^\circ) = 12.79 \text{ m/s} \]
03

Calculate Maximum Height Above Ground

Once the engines turn off, the rocket's vertical motion can be analyzed as uniformly accelerated motion under gravity. We use the kinematic equation:\[ v_{y,f}^2 = v_y^2 - 2g \cdot h \]where \(v_{y,f} = 0\) (at maximum height), \(v_y = 12.79 \text{ m/s}\), and \(g = 9.81 \text{ m/s}^2\). Solve for \(h\):\[ 0 = 12.79^2 - 2 \cdot 9.81 \cdot h \]\[ h = \frac{12.79^2}{2 \cdot 9.81} = 8.34 \text{ m} \]
04

Calculate Total Height Above Ground

The total height of the rocket above the ground includes both the height of the incline and the additional height gained after leaving the incline. The height of the incline is calculated using:\[ h_{incline} = 200 \cdot \sin(35^\circ) = 114.7 \text{ m} \]Thus, the total maximum height is:\[ h_{total} = 114.7 + 8.34 = 123.04 \text{ m} \]
05

Calculate Horizontal Range

The horizontal range beyond point A can be determined by calculating the time of flight and using horizontal velocity. The time to reach the maximum height is calculated by:\[ t_{up} = \frac{v_y}{g} = \frac{12.79}{9.81} = 1.30 \text{ s} \]The time to come down from the maximum height is the same, so the total time \( t_{total} = 2 \cdot t_{up} = 2.60 \text{ s} \).The horizontal range is:\[ R = v_x \cdot t_{total} = 18.30 \times 2.60 = 47.58 \text{ m} \]

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Key Concepts

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

Projectile Motion
Projectile motion refers to the motion of an object that is projected into the air and subject to the force of gravity. In the case of our rocket, it's launched up a 35-degree inclined plane before being influenced only by gravity. When the engines shut down, it follows a curved path. This type of motion can be understood by analyzing the independent vertical and horizontal components separately.
  • Horizontal motion remains uniform since there鈥檚 no air resistance, meaning the horizontal velocity is constant.
  • Vertical motion, conversely, is uniformly accelerated due to gravity.
Understanding projectile motion is vital for solving problems like predicting the maximum height and range of objects in motion. This motion follows a parabola, determined by initial speed and launch angle.
Incline Motion
Incline motion involves an object moving along a sloped surface. Here, our rocket ascends a 200-meter incline with an initial acceleration provided by its engines. This inclination makes the analysis of the movement slightly more complex, as the direction of the motion isn't purely horizontal or vertical.
  • The incline affects how we resolve the velocity, since the object is not moving along a simple horizontal or vertical path.
  • The rise of 35 degrees means that both the gravitational component along the incline and the acceleration component must be taken into account.
As the rocket moves up the incline, it gains velocity which it then uses to continue as a projectile when the engines turn off. The incline height is crucial for calculating the total maximum height reached.
Acceleration
Acceleration is the rate of change of velocity with respect to time. For our rocket, it starts with a constant acceleration of 1.25 m/s虏 up the incline. This acceleration is pivotal since it determines the final velocity of the rocket at the end of the incline, which initiates the projectile motion phase.
  • Acceleration allows the rocket to gain speed over the distance covered on the incline.
  • This initial acceleration phase ends the moment the rocket leaves the incline and only gravity acts upon it.
Understanding this concept is important for breaking down how speed is gained and predicting future motion once the rocket is airborne.
Velocity Components
The velocity of an object in motion can be separated into horizontal and vertical components, especially useful when dealing with inclined plane motion and projectile motion. For our rocket, after ascending the incline, it has velocity components determined by the 35-degree angle of the incline. By using trigonometric functions:
  • The horizontal component, \(v_x = v \cdot \cos(35^\circ)\), describes the motion parallel to the ground.
  • The vertical component, \(v_y = v \cdot \sin(35^\circ)\), describes the motion perpendicular to the ground.
These components help us determine the path and maximum height the rocket will achieve upon becoming a projectile. Analyzing velocity in components aids in calculating the horizontal range and ensuring complete understanding of projectile motions.

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Most popular questions from this chapter

A cart carrying a vertical missile launcher moves horizontally at a constant velocity of \(30.0 \mathrm{~m} / \mathrm{s}\) to the right (Figure 3.39 ). It launches a rocket vertically upward. The missile has an initial vertical velocity of \(40.0 \mathrm{~m} / \mathrm{s}\) relative to the cart. (a) How high does the rocket go? (b) How far does the cart travel while the rocket is in the air? (c) Where does the rocket land relative to the cart?

A physics book slides off a horizontal tabletop with a speed of \(1.10 \mathrm{~m} / \mathrm{s} .\) It strikes the floor in \(0.350 \mathrm{~s}\). Ignore air resistance. Find (a) the height of the tabletop above the floor, (b) the horizontal distance from the edge of the table to the point where the book strikes the floor, and (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.

A football is kicked from ground level at a speed of \(25 \mathrm{~m} / \mathrm{s}\). If it reaches a maximum height of \(24 \mathrm{~m},\) at what angle was it kicked (relative to horizontal)?

A firefighting crew uses a water cannon that shoots water at \(25.0 \mathrm{~m} / \mathrm{s}\) at a fixed angle of \(53.0^{\circ}\) above the horizontal. The firefighters want to direct the water at a blaze that is \(10.0 \mathrm{~m}\) above ground level. How far from the building should they position their cannon? There are \(t w o\) possibilities; can you get them both? (Hint: Start with a sketch showing the trajectory of the water.)

A player kicks a football at an angle of \(40.0^{\circ}\) from the horizontal, with an initial speed of \(12.0 \mathrm{~m} / \mathrm{s}\). A second player standing at a distance of \(30.0 \mathrm{~m}\) from the first (in the direction of the kick) starts running to meet the ball at the instant it is kicked. How fast must he run in order to catch the ball just before it hits the ground?
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