/*! 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 78 A plane flying horizontally at \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 78

A plane flying horizontally at \(100 \mathrm{~m} / \mathrm{s}\) releases an object which reaches the ground in \(10 \mathrm{~s}\). At what angle with horizontal it hits the ground? a. \(55^{\circ}\) b. \(45^{\circ}\) c. \(60^{\circ}\) d. \(75^{\circ}\)

Short Answer

Expert verified
The object hits the ground at approximately \(45^{\circ}\).

Step by step solution

01

Understand the Problem

We have an object released from a plane flying horizontally. The plane's speed is given as \(100 \mathrm{~m/s}\), and the object takes \(10 \mathrm{~s}\) to reach the ground. We need to calculate the angle at which the object hits the ground.
02

Determine Horizontal and Vertical Components

Since the plane is flying horizontally at \(100 \mathrm{~m/s}\), the horizontal velocity component of the object remains \(100 \mathrm{~m/s}\). The vertical velocity can be calculated using the formula \( v = u + at \) where \(u=0\) (since it starts falling from rest vertically), \( a = 9.8 \mathrm{~m/s}^2\) (acceleration due to gravity), and \( t = 10 \mathrm{~s}\). Thus, \( v_y = 0 + 9.8 \times 10 = 98 \mathrm{~m/s} \).
03

Calculate the Resultant Velocity

The resultant velocity \( v \) of the object is given by the Pythagorean theorem: \( v = \sqrt{v_x^2 + v_y^2} \), where \( v_x = 100 \mathrm{~m/s} \) and \( v_y = 98 \mathrm{~m/s} \). So, \( v = \sqrt{100^2 + 98^2} \).
04

Calculate the Angle with the Horizontal

The angle \( \theta \) can be calculated using the formula \( \tan \theta = \frac{v_y}{v_x} \). Therefore, \( \tan \theta = \frac{98}{100} = 0.98 \). The angle \( \theta \) is \( \tan^{-1}(0.98) \).
05

Find the Angle Using Arctangent Function

Using a calculator, find \( \theta = \tan^{-1}(0.98) \). This gives \( \theta \) approximately \( 44.4^{\circ} \), which rounds to \( 45^{\circ} \).

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 91影视!

Key Concepts

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

Horizontal Velocity
In projectile motion, horizontal velocity refers to the speed at which an object travels parallel to the ground. When an object is released from a moving plane, like in our example, it carries forward this horizontal speed.
The horizontal velocity is constant throughout the object's flight because, in the absence of air resistance, no net external force acts horizontally.
  • In our problem, the plane's horizontal speed is given as 100 m/s. This means the object continues moving forward at this same speed as it descends.
  • This constant velocity implies that the horizontal motion is independent of vertical motion, which abides by gravitational laws.
Understanding this concept is crucial because it helps us differentiate horizontal from vertical components in two-dimensional motion analysis.
Vertical Velocity
The vertical velocity of an object in projectile motion is influenced by gravity. This velocity starts at zero when an object is released horizontally, but it increases linearly over time due to gravitational acceleration.
In this exercise, we use the formula for velocity under constant acceleration:
In our scenario, the formula becomes:
  • \( v_y = 0 + (9.8 \text{ m/s}^2 \times 10 \text{ s}) = 98 \text{ m/s} \)
  • This calculation shows that the object acquires a vertical speed of 98 m/s downward by the time it hits the ground.
When solving problems involving vertical motion, always remember to factor in gravity's consistent acceleration, which in this case is approximately 9.8 m/s虏 on Earth.
Angle of Projection
The angle of projection determines the direction at which an object is launched or, in some cases, as we see here, hits the target. Given both the horizontal and vertical velocity components, you can determine this angle relative to the horizontal by using trigonometric functions like arctangent.
For our problem, we find the angle at which the object hits the ground using the formula:
  • \( \tan \theta = \frac{v_y}{v_x} \)
  • Plug in the numbers: \( \tan \theta = \frac{98}{100} = 0.98 \)
  • The angle, \( \theta \), can then be found as: \( \theta = \tan^{-1}(0.98) \)
This mathematical approach shows that the angle at which the object strikes the ground is approximately 45掳, which aligns with the answer choice given in the original exercise.
Understanding how to calculate this angle is vital in projectile problems because it ties together the motion's horizontal and vertical aspects through an intuitive geometric perspective.

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 small body is dropped from a rising balloon. A person \(A\) stands on ground, while another person \(B\) is on the balloon. Choose the correct statement: Immediately, after the body is released. a. \(A\) and \(B\), both feel that the body is coming (going) down. b. \(A\) and \(B\), both feel that the body is going up. c. \(A\) feels that the body is coming down, while \(B\) feels that the body is going up. d. \(A\) feels that the body is going up, while \(B\) feels that the body is going down.

i. If air resistance is not considered in a projectile motion, the horizontal motion takes place with a. constant velocity b. constant acceleration c. constant retardation d. variable velocityii. When a projectile is fired at an angle \(\theta\) ith horizontal with velocity \(u\), then its vertical component a. remains the sameb. goes on increasing with the height c. first decreases and then increases with the height d. first increases then decreases with the height iii. Range of a projectile is \(R\), when the angle of projection is \(30^{\circ}\). Then, the value of the other angle of projection for the same range is a. \(45^{\circ}\) b. \(60^{\circ}\) c. \(50^{\circ}\) d. \(40^{\circ}\) iv. A ball is thrown upwards and it returns to ground describing a parabolic path. Which of the following quantitics remains constant throughout the motion? a. kinetic energy of the ball b. speed of the ball c. horizontal component of velocity d. vertical component of velocity

A person takes an aim at a monkey sitting on a tree and fires a buliet. Seeking the smoke the monkey begins to fall freely; then the bullet will a. always hit the monkey b. go above the monkey c. go below the monkey d. hit the monkey if the initial velocity of the bullet is\\} more than a certain velocity

A man is walking on a road with a velocity \(3 \mathrm{kmh}^{-1}\). Suddenly rain starts falling. The velocity of rain is \(10 \mathrm{kmh}^{-1}\) in vertically downward direction. The relative velocity of the rain wr.t. man is a. \(\sqrt{7} \mathrm{kmh}^{-1}\) b. \(\sqrt{13} \mathrm{kmh}^{-1}\) c. \(13 \mathrm{kmh}^{-1}\) d. \(\sqrt{109} \mathrm{kmh}^{-1}\)

A man can swim in still water with a speed of \(2 \mathrm{~m} / \mathrm{s}\). If he wants to cross a river of water current speed \(\sqrt{3} \mathrm{~m} / \mathrm{s}\) along the shortest possible path, then in which direction should he swim? a. At an angle \(120^{\circ}\) to the water current. b. At an angle \(150^{\circ}\) to the water current. c. At an angle \(90^{\circ}\) to the water current. d. None of these
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Kinematics Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Quantum Physics

Read Explanation

Space 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.