Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 51

You are working as a consultant on a video game designing a bomb site for a World War I airplane. In this game, the plane you are flying is traveling horizontally at \(40.0 \mathrm{m} / \mathrm{s}\) at an altitude of $125 \mathrm{m}$ when it drops a bomb. (a) Determine how far horizontally from the target you should release the bomb. (b) What direction is the bomb moving just before it hits the target?

Short Answer

Expert verified
Answer: The horizontal distance to the target is approximately 202 meters, and the bomb is moving at an angle of approximately 51.2 degrees below the horizontal just before it hits the ground.

Step by step solution

01

(Step 1: List Known Values)

Known values are: - Horizontal velocity of the plane, \(v_x = 40 m/s\) - Altitude of the plane, \(h = 125 m\) - Acceleration due to gravity, \(g = 9.81 m/s^2\)
02

(Step 2: Calculate the Time of Flight)

First, we need to find out how long it takes for the bomb to reach the ground, which is also called time of flight. From the equation \(y = v_{0y}t - \frac{1}{2}gt^2\), we have: $$ h = 0 \cdot t - \frac{1}{2} \cdot g \cdot t^2 $$ Solving for \(t\): $$ t = \sqrt{\frac{2h}{g}} \text{, as } v_{0y} = 0 $$ Plugging in the known values: $$ t = \sqrt{\frac{2 \cdot 125}{9.81}} \approx 5.05 s $$ So, the time of flight for the bomb is \(t \approx 5.05 s\).
03

(Step 3: Calculate the Horizontal Distance)

Now, we can find the horizontal distance \(x\) by using the following equation: $$ x = v_x \cdot t $$ Plugging the known values: $$ x = 40 \cdot 5.05 \approx 202 m $$ Thus, the horizontal distance to the target is approximately \(202 m\).
04

(Step 4: Calculate the Final Vertical Velocity)

To find the final vertical velocity \(v_{y}\) just before the bomb hits the ground, we will use the equation: $$ v_{y} = v_{0y} - g \cdot t $$ Plugging the values: $$ v_{y} = 0 - 9.81 \cdot 5.05 \approx -49.6\ m/s $$
05

(Step 5: Calculate the Direction of the Bomb)

Now that we have both horizontal and vertical components of the velocity, we can calculate the angle \(\theta\) using the following formula: $$ \theta = \arctan{\frac{-v_{y}}{v_x}} $$ Plugging the values: $$ \theta = \arctan{\frac{49.6}{40}} \approx 51.2^\circ $$ The answer to (a) is that the bomb should be released approximately \(202 m\) horizontally from the target. For (b), the bomb is moving at an angle of approximately \(51.2^\circ\) below the horizontal just before it hits the target.

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

Two angles are complementary when their sum is \(90.0^{\circ}\) Find the ranges for two projectiles launched with identical initial speeds of $36.2 \mathrm{m} / \mathrm{s}$ at angles of elevation above the horizontal that are complementary pairs. (a) For one trial, the angles of elevation are \(36.0^{\circ}\) and \(54.0^{\circ} .\) (b) For the second trial, the angles of elevation are \(23.0^{\circ}\) and \(67.0^{\circ} .\) (c) Finally, the angles of elevation are both set to \(45.0^{\circ} .\) (d) What do you notice about the range values for each complementary pair of angles? At which of these angles was the range greatest?
A car travels east at \(96 \mathrm{km} / \mathrm{h}\) for \(1.0 \mathrm{h}\). It then travels \(30.0^{\circ}\) east of north at \(128 \mathrm{km} / \mathrm{h}\) for \(1.0 \mathrm{h} .\) (a) What is the average speed for the trip? (b) What is the average velocity for the trip?
A runner is practicing on a circular track that is \(300 \mathrm{m}\) in circumference. From the point farthest to the west on the track, he starts off running due north and follows the track as it curves around toward the east. (a) If he runs halfway around the track and stops at the farthest eastern point of the track, what is the distance he traveled? (b) What is his displacement?
A vector \(\overrightarrow{\mathbf{A}}\) has a magnitude of \(22.2 \mathrm{cm}\) and makes an angle of \(130.0^{\circ}\) with the positive \(x\) -axis. What are the \(x\) - and \(y\) -components of this vector?
A boat that can travel at \(4.0 \mathrm{km} / \mathrm{h}\) in still water crosses a river with a current of \(1.8 \mathrm{km} / \mathrm{h}\). At what angle must the boat be pointed upstream to travel straight across the river? In other words, in what direction is the velocity of the boat relative to the water?
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Radiation

Read Explanation

Nuclear Physics

Read Explanation

Scientific Method Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Medical Physics

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