Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 93

A beanbag is thrown horizontally from a dorm room window a height \(h\) above the ground. It hits the ground a horizontal distance \(h\) (the same distance \(h\) ) from the dorm directly below the window from which it was thrown. Ignoring air resistance, find the direction of the beanbag's velocity just before impact.

Short Answer

Expert verified
Answer: The direction of the beanbag's velocity just before impact is given by the angle, \(\theta\), which can be calculated using the following equation: $$ \theta = \tan^{-1}\left(\frac{\sqrt{2gh}}{\sqrt{\frac{gh}{2}}}\right) $$ Here, \(g\) is the gravitational acceleration (approximately \(9.81\,\text{m/s}^2\)), and \(h\) is the height from which the beanbag is thrown.

Step by step solution

01

Identify the given information

The beanbag is thrown horizontally from a height, \(h\), above the ground. It hits the ground a horizontal distance, \(h\) from the dorm directly below the window from which it was thrown. We need to find the direction of the beanbag's velocity just before impact.
02

Apply the equations of motion

First, let's consider the horizontal motion of the beanbag. Since there is no air resistance, the horizontal velocity, \(v_x\), remains constant. Now, let's look at the vertical motion of the beanbag. It is initially at rest and then falls freely under the influence of gravity. We can use the following equation for the vertical motion: $$ y = h - \frac{1}{2}gt^2 $$ Where \(y\) is the distance traveled in the vertical direction, \(g\) is the gravitational acceleration (approximately \(9.81\,\text{m/s}^2\)), and \(t\) is the time taken for the beanbag to travel from the window to the ground.
03

Find the time of flight

First, we need to find the time it takes for the beanbag to hit the ground. We can use the vertical motion equation and solve for \(t\): $$ h = \frac{1}{2}gt^2 $$ Rearrange the equation to find the time, \(t\): $$ t = \sqrt{\frac{2h}{g}} $$
04

Determine the horizontal velocity

Now that we have the time of flight, we can find the horizontal velocity, \(v_x\). The horizontal distance is equal to the product of the horizontal velocity and the time of flight: $$ h = v_x t $$ Substituting the time of flight equation from Step 3: $$ h = v_x \sqrt{\frac{2h}{g}} $$ Now, solve for the horizontal velocity, \(v_x\): $$ v_x = \sqrt{\frac{gh}{2}} $$
05

Calculate the vertical velocity just before impact

Next, we need to find the vertical velocity, \(v_y\), just before the beanbag hits the ground. We can use the following equation for the vertical velocity: $$ v_y = gt $$ Substituting the time of flight equation from Step 3: $$ v_y = g\sqrt{\frac{2h}{g}} $$ $$ v_y = \sqrt{2gh} $$
06

Determine the direction of the velocity just before impact

Now, we can find the direction of the velocity just before impact by calculating the angle, \(\theta\), between the horizontal and the velocity vector using the horizontal and vertical velocities: $$ \tan\theta = \frac{v_y}{v_x} $$ Substituting the values of \(v_x\) and \(v_y\) calculated in steps 4 and 5: $$ \tan\theta = \frac{\sqrt{2gh}}{\sqrt{\frac{gh}{2}}} $$ Now, finding the value of the angle, \(\theta\): $$ \theta = \tan^{-1}\left(\frac{\sqrt{2gh}}{\sqrt{\frac{gh}{2}}}\right) $$ The angle, \(\theta\), represents the direction of the beanbag's velocity just before impact.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle's constant acceleration is north at $100 \mathrm{m} / \mathrm{s}^{2}\( At \)t=0,\( its velocity vector is \)60 \mathrm{m} / \mathrm{s}$ east. At what time will the magnitude of the velocity be $100 \mathrm{m} / \mathrm{s} ?$
A helicopter is flying horizontally at \(8.0 \mathrm{m} / \mathrm{s}\) and an altitude of \(18 \mathrm{m}\) when a package of emergency medical supplies is ejected horizontally backward with a speed of \(12 \mathrm{m} / \mathrm{s}\) relative to the helicopter. Ignoring air resistance, what is the horizontal distance between the package and the helicopter when the package hits the ground?
A boy is attempting to swim directly across a river; he is able to swim at a speed of \(0.500 \mathrm{m} / \mathrm{s}\) relative to the water. The river is \(25.0 \mathrm{m}\) wide and the boy ends up at \(50.0 \mathrm{m}\) downstream from his starting point. (a) How fast is the current flowing in the river? (b) What is the speed of the boy relative to a friend standing on the riverbank?
A speedboat heads west at \(108 \mathrm{km} / \mathrm{h}\) for 20.0 min. It then travels at \(60.0^{\circ}\) south of west at \(90.0 \mathrm{km} / \mathrm{h}\) for 10.0 min. (a) What is the average speed for the trip? (b) What is the average velocity for the trip?
A projectile is launched at \(t=0\) with initial speed \(v_{\mathrm{i}}\) at an angle \(\theta\) above the horizontal. (a) What are \(v_{x}\) and \(v_{y}\) at the projectile's highest point? (b) Find the time \(t\) at which the projectile reaches its maximum height. (c) Show that the maximum height \(H\) of the projectile is $$H=\frac{\left(v_{i} \sin \theta\right)^{2}}{2 g}$$
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Work Energy and Power

Read Explanation

Kinematics Physics

Read Explanation

Space Physics

Read Explanation

Classical Mechanics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free