/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A pistol that fires a signal fla... [FREE SOLUTION] | 91Ó°ÊÓ

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A pistol that fires a signal flare gives it an initial velocity (muzzle velocity) of 125 \(\mathrm{m} / \mathrm{s}\) at an angle of \(55.0^{\circ}\) above the borizontal. You can ignore air resistance. Find the flare's maximum height and the distance from its firing point to its landing point if it is fired (a) on the level salt flats of Utah, and (b) over the flat Sea of Tranquility on the Moon, where \(g=1.67 \mathrm{m} / \mathrm{s}^{2}\) .

Short Answer

Expert verified
Earth: max height ~750 m, range ~1800 m. Moon: max height ~4400 m, range ~10600 m.

Step by step solution

01

Define Known Values and Split the Initial Velocity into Components

The initial velocity \(v_0\) is 125 m/s and the launch angle \(\theta\) is 55.0°. The acceleration due to gravity \(g\) on Earth is 9.81 m/s². We first need to split the initial velocity into horizontal and vertical components:\[ v_{0x} = v_0 \cos \theta = 125 \cos 55.0°\]\[ v_{0y} = v_0 \sin \theta = 125 \sin 55.0°\]Calculate these components.
02

Calculate the Maximum Height on Earth

To find the maximum height, we use:\[ v_{y}^2 = v_{0y}^2 - 2g h_{max}\]At maximum height, the vertical velocity \(v_{y}\) is 0:\[ 0 = (v_0 \sin 55.0°)^2 - 2 \times 9.81 \times h_{max}\]Solve for \(h_{max}\) on Earth.
03

Calculate Time of Flight on Earth

The time to reach the maximum height \(t_{up}\) can be calculated using:\[ t_{up} = \frac{v_{0y}}{g} = \frac{v_0 \sin 55.0°}{9.81}\]The total time of flight \(T\) is twice \(t_{up}\):\[ T = 2t_{up}\]Calculate \(T\).
04

Calculate the Horizontal Range on Earth

The horizontal range \(R\) is calculated using the horizontal velocity and the total time of flight:\[ R = v_{0x} \times T = v_0 \cos 55.0° \times T\]Find the range \(R\) on Earth.
05

Calculate Maximum Height and Range on the Moon

On the Moon, where \(g = 1.67 \mathrm{m/s}^2\), repeat the processes from Steps 2-4:1. Use the Moon's gravity to find the maximum height \(h_{max}\): \[ h_{max} = \frac{(v_0 \sin 55.0°)^2}{2 \times 1.67} \]2. Time of flight \(T\) becomes: \[ T = 2 \times \frac{v_0 \sin 55.0°}{1.67} \]3. Calculate the horizontal range \(R\): \[ R = v_0 \cos 55.0° \times T \]Solve these to find the maximum height and range on the Moon.

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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 fundamental in physics when dealing with motion. They allow us to predict future positions and velocities of an object in motion, given initial conditions like initial speed, angle, and acceleration. These equations serve as the foundation for solving projectile motion problems. They help relate time, velocity, acceleration, and displacement: all crucial elements for analyzing any object launched through a projectile motion.
Imagine an object being thrown into the air. The kinematic equations let you systematically analyze every phase of its flight. This includes the total time it stays airborne, how high it gets, how far it travels horizontally, and its instantaneous speeds at every point in its trajectory.
  • The main kinematic equation to find the vertical position is: \[ v_y^2 = v_{0y}^2 - 2 g h_{max} \]
  • For time calculations: \[ t = \frac{v_{0y}}{g} \]
  • To evaluate the range: \[ R = v_{0x} \cdot T \]
These equations bridge the gaps in visualizing a projectile's path from its start to end, allowing you to calculate the maximum height and horizontal distance covered.
Gravity on Earth and Moon
Gravity is the force that pulls objects towards the center of a celestial body, like Earth or the Moon. Its magnitude can vary drastically, altering how projectiles behave on different surfaces. On Earth, the gravitational pull is approximately 9.81 m/s². This makes any projectile follow a curved trajectory, with its vertical component directly affected by Earth's gravity causing it to eventually fall back down.
On the Moon, gravity is weaker, only about 1.67 m/s². This lesser gravity means that a projectile launched with the same initial speed and angle as on Earth would stay in the air longer and travel farther before landing.
The key takeaway is:
  • Earth: Gravity = 9.81 m/s²
  • Moon: Gravity = 1.67 m/s²
The Moon’s weaker gravity means less pull downward, resulting in different outcomes for maximum height and range when compared with Earth. Understanding these differences is crucial in predicting projectile motion accurately in varying gravitational conditions.
Components of Velocity
In projectile motion, the initial velocity can be divided into horizontal and vertical components. This division is essential to understanding how the projectile will move through the air. For example, if an object is launched with a certain initial speed, the horizontal component defines how quickly it moves across a surface, while the vertical component influences how high it will travel.
  • The horizontal component (\( v_{0x} \)) uses the cosine of the launch angle: \[ v_{0x} = v_0 \cos \theta \]
  • The vertical component (\( v_{0y} \)) uses the sine of the angle: \[ v_{0y} = v_0 \sin \theta \]
These calculations ensure we separate motion into its directional components. With this, we can address vertical movements (like finding maximum height) and horizontal progressions (like determining the range) more effectively. Think about throwing a ball: by looking at the velocity components, we can predict not just *where* it will land but *how* it travels to get there.
Maximum Height Calculation
The maximum height a projectile reaches is a critical factor in understanding its motion. It represents the peak of the object's vertical trajectory before gravity takes over and begins to pull it back down. Calculating this involves considering the vertical component of the initial velocity and the gravitational force acting on the projectile. At its maximum height, the vertical velocity becomes zero because the projectile momentarily stops climbing.
To find this height, you use: \[ v_y^2 = v_{0y}^2 - 2g h_{max} \]
For Earth, where gravity is 9.81 m/s², this formula helps us determine how high the projectile will climb before descending. On the Moon, where gravity is lighter at 1.67 m/s², the projectile would reach a higher max height given the same initial conditions, simply because there’s less downward pull.
Knowing the maximum height is vital when evaluating a projectile's overall path and performance, especially in environments with different gravitational forces.

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Most popular questions from this chapter

A railroad flatear is traveling to the right at a speed of 13.0 \(\mathrm{m} / \mathrm{s}\) relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (Fig. 3.43\()\) . What is the velocity (magnitude and direction) of the motor scooter relative to the flatcar if its velocity relative to the observer on the ground is (a) 18.0 \(\mathrm{m} / \mathrm{s}\) to the right? (b) 3.0 \(\mathrm{m} / \mathrm{s}\) to the left? (c) zero?

A rocket designed to place small payloads into orbit is carricd to an altitude of 12.0 \(\mathrm{km}\) above sca level by a converted air- liner. When the airtiner is flying in a straight line at a constant speed of 850 \(\mathrm{km} / \mathrm{h}\) , the rocket is dropped. After the drop, the airliner maintains the same altitude and speed and continues to fly in a straight line. The rocket falls for a brief time, after which its rocket motor turms on. Once its rocket motor is on, the combined effects of thrust and gravity give the rocket a constant acceleration of magnitude 3.00 \(\mathrm{g}\) directed at an angle of \(30.0^{\circ}\) above the horizontal. For reasons of safety, the rocket should be at least 1.00 \(\mathrm{km}\) in front of the airliner when it climbs through the airliner's altitade. Your job is to determine the minimum time that the rocket must fall before its engine starts. You can ignore air resistance.

Tossing Your Lunch. Henrietta is going off to her physics class, jogging down the sidewalk at 3.05 \(\mathrm{m} / \mathrm{s}\) . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is 43.9 \(\mathrm{m}\) above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally 9.00 \(\mathrm{s}\) after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance. (a) With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? (b) Where is Henrietta when she catches the bagels?

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City A lies directly west of city \(B\) . When there is no wind, an airliner makes the 5550 -kmround-trip fight between them in 6.60 \(\mathrm{h}\) of flying time while traveling at the same speed in both directions. When a strong, steady \(225-\mathrm{km} / \mathrm{h}\) wind is blowing from west to east and the airliner has the same airspeed as before, how long will the trip take?

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