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Chapter 3: Problem 67

A cart carrying a vertical missile launcher moves horizontally at a constant velocity of 30.0 m/s to the right. It launches a rocket vertically upward. The missile has an initial vertical velocity of 40.0 m/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?

Short Answer

Expert verified
(a) 81.6 m; (b) 244.8 m; (c) Rocket lands directly above the launch point relative to the cart.

Step by step solution

01

Identify the problem for Part (a)

To find the maximum height reached by the rocket, we evaluate the vertical motion. The rocket is in free-fall with an initial vertical velocity of 40.0 m/s and an acceleration due to gravity of \( g = -9.8 \text{ m/s}^2 \). The maximum height is reached when the vertical velocity becomes 0 m/s.
02

Calculate the time to reach maximum height

We use the kinematic equation for vertical motion: \( v = u + at \),where \( v = 0 \), \( u = 40.0 \text{ m/s} \), and \( a = -9.8 \text{ m/s}^2 \). Solving for \( t \), \[ 0 = 40.0 \text{ m/s} - 9.8 \text{ m/s}^2 \times t \]\( t = \frac{40.0 \text{ m/s}}{9.8 \text{ m/s}^2} \approx 4.08 \text{ s} \).
03

Calculate maximum height

Using the time to reach maximum height, calculate the maximum height with the formula:\( h = ut + \frac{1}{2}at^2 \),substituting in the values:\[ h = 40.0 \text{ m/s} \times 4.08 \text{ s} + \frac{1}{2} \times (-9.8 \text{ m/s}^2) \times (4.08 \text{ s})^2 \approx 81.6 \text{ m} \].
04

Calculate total time in air

The total time in the air is double the time to reach the maximum height since the ascent time equals the descent time. Thus, \( t_{total} = 2 \times 4.08 \text{ s} \approx 8.16 \text{ s} \).
05

Determine distance cart travels (Part b)

The cart travels horizontally at 30.0 m/s for the duration the rocket is in the air. Using the total time:\( d = v \times t_{total} \ = 30.0 \text{ m/s} \times 8.16 \text{ s} = 244.8 \text{ m} \).
06

Relative landing position of the rocket (Part c)

The horizontal component of the rocket's velocity equals the cart's constant speed (30.0 m/s). Consequently, the rocket moves horizontally at the same speed as the cart throughout its journey. Thus, the rocket lands directly above where it was launched, relative to the cart.

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Key Concepts

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

Kinematic Equations
Kinematic equations are essential tools for solving problems in physics related to motion. These equations describe the relationships between velocity, acceleration, displacement, and time. They are particularly useful in projectile motion, where objects move under the influence of gravity.

In our problem, the kinematic equations help us determine how the rocket travels vertically. The specific equation used here is: \[ v = u + at \] where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

This equation allowed us to find the time it takes for the rocket to reach its maximum height by setting the final velocity (\( v \) ) to zero because at the highest point of vertical motion, the upward speed is zero before it falls back down.
Vertical Motion
Vertical motion focuses on how the rocket moves up and down under the influence of gravity. When discussing vertical motion, it's important to remember that gravity is the only force affecting the rocket's movement after it's launched. This is crucial in determining both the rocket's maximum height and the time it takes to reach it.

The initial vertical velocity given (40.0 m/s) and the constant acceleration due to gravity (\(-9.8 \text{ m/s}^2\)) help us map the rocket's journey in the vertical direction. Using the kinematic equation, we calculated how long it takes the rocket to stop ascending (4.08 seconds).

Once it reaches that point, the rocket begins to descend, and the time it takes to go down is the same, resulting in a total flight time of 8.16 seconds. This information is pivotal when calculating the maximum height reached (81.6 meters), using the equation: \[ h = ut + \frac{1}{2}at^2 \] which considers both the initial velocity and the effect of gravity on the rocket's ascent.
Constant Velocity
Constant velocity plays a key role in the horizontal motion of the cart. While the rocket moves vertically, the cart continues to travel horizontally at a uniform speed of 30.0 m/s.

In physics, constant velocity implies that the speed and direction of an object remain unchanged over time. This characteristic simplifies calculations since the horizontal distance can be found by simply multiplying the velocity by the total time the rocket is in the air.

Thus, the distance travelled by the cart during the rocket's flight is:\[ d = v \times t_{total} = 30.0 \text{ m/s} \times 8.16 \text{ s} = 244.8 \text{ m} \]This shows that while the rocket is in vertical motion, the cart keeps moving steadily along its path.

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Most popular questions from this chapter

A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0\(^{\circ}\) above the horizontal toward the cliff. (a) What must the minimum muzzle velocity be for the shell to clear the top of the cliff? (b) The ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon. Under the conditions of part (a), how far does the shell land past the edge of the cliff?

Firemen use a high-pressure hose to shoot a stream of water at a burning building. The water has a speed of 25.0 m/s as it leaves the end of the hose and then exhibits projectile motion. The firemen adjust the angle of elevation \(\alpha\) of the hose until the water takes 3.00 s to reach a building 45.0 m away. Ignore air resistance; assume that the end of the hose is at ground level. (a) Find \(\alpha\). (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

A movie stuntwoman drops from a helicopter that is 30.0 m above the ground and moving with a constant velocity whose components are 10.0 m/s upward and 15.0 m/s horizontal and toward the south. Ignore air resistance. (a) Where on the ground (relative to the position of the helicopter when she drops) should the stuntwoman have placed foam mats to break her fall? (b) Draw \(x\)-\(t\), \(y\)-\(t\), \(v$$_x\)-\(t\), and \(v$$_y\)-\(t\) graphs of her motion.

The earth has a radius of 6380 km and turns around once on its axis in 24 h. (a) What is the radial acceleration of an object at the earth's equator? Give your answer in m/s\(^2\) and as a fraction of \(g\). (b) If \(a_{rad}\) at the equator is greater than \(g\), objects will fly off the earth's surface and into space. (We will see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determine the width of the river and the current speed (the speed of the water relative to the earth). You have a small boat with an outboard motor. By measuring the time it takes to cross a pond where the water isn't flowing, you have calibrated the throttle settings to the speed of the boat in still water. You set the throttle so that the speed of the boat relative to the river is a constant 6.00 m/s. Traveling due north across the river, you reach the opposite bank in 20.1 s. For the return trip, you change the throttle setting so that the speed of the boat relative to the water is 9.00 m/s. You travel due south from one bank to the other and cross the river in 11.2 s. (a) How wide is the river, and what is the current speed? (b) With the throttle set so that the speed of the boat relative to the water is 6.00 m/s, what is the shortest time in which you could cross the river, and where on the far bank would you land?
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