Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An airplane flies horizontally above the flat surface of a desert at an altitude of \(5.00 \mathrm{~km}\) and a speed of \(1000 . \mathrm{km} / \mathrm{h}\). If the airplane is to drop a bomb that is supposed to hit a target on the ground, where should the plane be with respect to the target when the bomb is released? If the target covers a circular area with a diameter of \(50.0 \mathrm{~m}\), what is the "window of opportunity" (or margin of error allowed) for the release time?

Short Answer

Expert verified
Answer: The position of the airplane relative to the target when the bomb should be released is 8870.62 meters away in the horizontal direction. The window of opportunity for the bomb release is 0.20 seconds.

Step by step solution

01

Find the time it takes for the bomb to hit the ground

First, we need to find the time it will take for the bomb to fall from the airplane to the ground. We can use the equation of motion for free-falling objects with constant gravitational acceleration: \(h = \frac{1}{2}gt^2\) where \(h = 5.00 \ km\) (convert to meters), \(g = 9.81 \ m/s^{2}\) (acceleration due to gravity), and \(t\) is the time it takes for the bomb to hit the ground. Rearrange the equation, and solve for \(t\): \(t = \sqrt{\frac{2h}{g}}\) Plug in the given values: \(t = \sqrt{\frac{2(5000)}{9.81}} = 31.9 \ \text{seconds}\)
02

Find the horizontal distance traveled by the bomb

Now that we have the time it takes for the bomb to reach the ground, we can find the horizontal distance traveled by the bomb. We know the speed of the airplane, which is also the horizontal speed of the bomb, and it remains constant throughout the fall. The horizontal distance is the product of horizontal speed and time. \(v_x = 1000 \ \frac{\mathrm{km}}{\mathrm{hour}}\) (convert to \(\frac{\mathrm{m}}{\mathrm{s}}\)) \(v_x = 1000 \cdot \frac{1000}{3600} = 277.8 \ \frac{\mathrm{m}}{\mathrm{s}}\) Now, find the horizontal distance: \(x = v_x \cdot t = 277.8 \cdot 31.9 = 8870.62 \ \text{m}\)
03

Calculate the position of the airplane relative to the target

The position of the airplane relative to the target when the bomb should be released is: \(x = 8870.62 \ \text{m}\)
04

Calculate the window of opportunity for the bomb release

To find the window of opportunity, we need to find the difference in time between earliest and latest release times. We're given the target area's diameter, so we can use half of the diameter as the margin of error for both early and late release times. Half of the target diameter is \(25 \ \text{m}\). Add this to our horizontal distance from the airplane to the target: \(x_1 = 8870.62 + 25 = 8895.62 \ \text{m}\) Subtract the margin of error from the horizontal distance: \(x_2 = 8870.62 - 25 = 8845.62 \ \text{m}\) Now find the new times \(t_1\) and \(t_2\): \(t_1 = \frac{x_1}{v_x} = \frac{8895.62}{277.8} \approx 32.02 \ \text{seconds}\) \(t_2 = \frac{x_2}{v_x} = \frac{8845.62}{277.8} \approx 31.82 \ \text{seconds}\) The window of opportunity (margin of error) is the difference between the two times: \(window \ of \ opportunity = t_1 - t_2 = 32.02 - 31.82 = 0.20 \ \text{seconds}\) So, the airplane should be \(8870.62 \ \text{m}\) away from the target in the horizontal direction when the bomb is released, and the window of opportunity for the bomb release is \(0.20 \ \text{seconds}\).

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.

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

Key Concepts

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

Free Fall and Acceleration Due to Gravity
Imagine dropping an apple from a tree; it falls straight to the ground without any horizontal movement. This is a typical example of free fall, a concept crucial in understanding projectile motion. Free fall occurs when the only force acting on an object is gravity, causing it to accelerate towards the Earth constantly. That constant acceleration, known as acceleration due to gravity, is denoted by the symbol g and has an average value of about \(9.81 \ m/s^2\) on Earth's surface.

When an object is in free fall, its vertical motion can be described by kinematic equations, which take into consideration the acceleration due to gravity. These equations allow us to predict how long an object will take to hit the ground (time of fall), how fast it will be moving when it lands (velocity upon impact), and how high the object will rise or fall (displacement). For instance, the equation \( h = \frac{1}{2}gt^2 \) calculates the distance (\(h\)) fallen after a certain amount of time (\(t\)), given the acceleration due to gravity (\(g\)).

In our exercise example, we used this concept to determine how long it will take for a bomb to travel from the airplane to the desert surface. This knowledge is paramount for the pilot to estimate where to release the payload to ensure it hits the target.
Kinematic Equations
Kinematic equations are the backbone of classical mechanics and are particularly useful in describing the motion of objects in terms of displacement, velocity, acceleration, and time without the influence of force or mass. In essence, these equations tell us how objects move. They become incredibly helpful in scenarios such as our exercise with the airplane and the bomb, especially since we are dealing with an object moving under the constant acceleration of gravity.

The general kinematic equations are:
  • \(v = v_0 + at\)
  • \(s = v_0t + \frac{1}{2}at^2\)
  • \(s = vt - \frac{1}{2}at^2\)
  • \(v^2 = v_0^2 + 2as\)

For an object in free fall, we often use the simplified form without initial velocity (\(v_0 = 0\)), because the object is starting from rest when dropped. Our exercise initially called for calculating the time it takes a bomb to reach the ground (\(t\)), which we found using the equation for vertical distance under constant acceleration due to gravity.
Motion in Two Dimensions
Motion in two dimensions, also known as projectile motion, involves the movement of an object both horizontally and vertically under the influence of gravity. Picture a soccer ball being kicked in an arc-shaped path; while gravity pulls it down, the ball still moves forward. Similarly, a dropped bomb will keep moving horizontally while falling towards earth due to gravity.

The key to solving problems involving two-dimensional motion is to treat the horizontal and vertical components separately. The horizontal motion is at a constant velocity (as there is no acceleration if we ignore air resistance), while the vertical motion has constant acceleration due to gravity. Using kinematic equations, we can analyze each component independently. In the horizontal direction, the primary equation used is distance travelled (\(x\)) equals velocity (\(v_x\)) times time (\(t\)): \(x = v_x \times t\).

In our plane-and-bomb problem, we used the constant horizontal speed of the plane to calculate how far the bomb will travel horizontally (\(x\)). Moreover, we illustrated the concept of a window of opportunity for bomb release, which is directly related to the precision required in such a two-dimensional movement scenario, ensuring that the bomb lands within the designated target area.

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

For a science fair competition, a group of high school students build a kicker-machine that can launch a golf ball from the origin with a velocity of \(11.2 \mathrm{~m} / \mathrm{s}\) and a launch angle of \(31.5^{\circ}\) with respect to the horizontal. a) Where will the golf ball fall back to the ground? b) How high will it be at the highest point of its trajectory? c) What is the ball's velocity vector (in Cartesian components) at the highest point of its trajectory? d) What is the ball's acceleration vector (in Cartesian components) at the highest point of its trajectory?

A cruise ship moves southward in still water at a speed of \(20.0 \mathrm{~km} / \mathrm{h}\) while a passenger on the deck of the ship walks toward the east at a speed of \(5.0 \mathrm{~km} / \mathrm{h}\). The passenger's velocity with respect to Earth is a) \(20.6 \mathrm{~km} / \mathrm{h}\), at an angle of \(14.04^{\circ}\) east of south. b) \(20.6 \mathrm{~km} / \mathrm{h}\), at an angle of \(14.04^{\circ}\) south of east. c) \(25.0 \mathrm{~km} / \mathrm{h}\), south. d) \(25.0 \mathrm{~km} / \mathrm{h},\) east. e) \(20.6 \mathrm{~km} / \mathrm{h}\), south.

A projectile is launched at an angle of \(45.0^{\circ}\) above the horizontal. What is the ratio of its horizontal range to its maximum height? How does the answer change if the initial speed of the projectile is doubled?

On a battlefield, a cannon fires a cannonball up a slope, from ground level, with an initial velocity \(v_{0}\) at an angle \(\theta_{0}\) above the horizontal. The ground itself makes an angle \(\alpha\) above the horizontal \(\left(\alpha<\theta_{0}\right)\). What is the range \(R\) of the cannonball, measured along the inclined ground? Compare your result with the equation for the range on horizontal ground (equation 3.25\()\)

During a jaunt on your sailboat, you sail \(2.00 \mathrm{~km}\) east, then \(4.00 \mathrm{~km}\) southeast, and finally an additional distance in an unknown direction. Your final position is \(6.00 \mathrm{~km}\) directly east of the starting point. Find the magnitude and direction of the third leg of your journey.

See all solutions

Recommended explanations on Physics Textbooks

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