/*! 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 21 Airplane An airplane, diving at ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 21

Airplane An airplane, diving at an angle of \(53.0^{\circ}\) with the vertical, releases a projectile at an altitude of \(730 \mathrm{~m}\). The projectile hits the ground \(5.00\) s after being released. (a) What is the speed of the aircraft? (b) How far did the projectile travel horizontally during its flight? What were the (c) horizontal and (d) vertical components of its velocity just before striking the ground?

Short Answer

Expert verified
(a) Solve for \(\text{v}\text{y}\); use \(\text{sin}(53.0^{\text{0}})\) to find \(\text{v}\text{i}\). (b) Find horizontal distance with \(\text{v}\text{i} \text{cos} (53.0^{\text{0}}) \text{t}\). (c) \(\text{v}\text{i}\text{cos} (53.0^{\text{0}})\). (d) \(\text{v}\text{i}\text{y} + \text{g}\text{t}\).

Step by step solution

01

- Analyze the given information

The projectile is released from an altitude of 730 m and hits the ground after 5.00 s. The diving angle is 53.0 degrees with the vertical.
02

- Determine vertical displacement and initial vertical velocity

The vertical displacement \(\text{y}\text{f} - \text{y}\text{i}\) can be found using the equation: \(\text{y}\text{f} = \text{y}\text{i} + \text{v}\text{i}\text{y}t - \frac{1}{2}\text{g}\text{t}^{2}\). Given that \(\text{y}\text{f} = 0 m\), \(\text{y}\text{i} = 730 m\), \(\text{t} = 5.00 s\), solve for initial vertical velocity \(\text{v}\text{i}\text{y}\).
03

- Simplify vertical motion equation

Rearrange \(\text{0} = 730 m + \text{v}\text{i}\text{y}(5.00 s) - \frac{1}{2}\text{(9.8 m/s}^{2})(5.00 s)^{2}\) to solve for \(\text{v}\text{i}\text{y}\).
04

- Solve for initial vertical velocity

Calculate \(\text{v}\text{i}\text{y} = \frac{1}{2} (9.8 m/s^{2}) ( 5.00 s)^{2} - 730 m = - \text{v}\text{i}\text{y} (5.00 s)\). The resulting \(\text{v}\text{i}\text{y}\) is negative, representing downward motion.
05

- Use trigonometric relations to find initial speed

\(\text{v}\text{y} = \text{v}\text{i} \text{sin}(53.0^{\text{0}})\). Given \(\text{v}\text{y}\), solve for \(\text{v}\text{i}\) which is the speed of the aircraft.
06

- Calculate initial speed

Use \(\text{v}\text{i} = \frac{\text{v}\text{y}}{\text{sin}(53.0^{\text{0}})}\) to find \(\text{v}\text{i}\).
07

- Calculate horizontal distance

The horizontal velocity component is \(\text{v}\text{i} \text{cos} (53.0^{\text{0}})\). Multiply by time \(\text{t} = 5.00 s\) to obtain horizontal distance.
08

- Calculate final horizontal and vertical velocities

Use projectile motion equations: final horizontal velocity \(\text{v}\text{f}\text{x}\) remains constant, \(\text{v}\text{i}\text{cos} (53.0^{\text{0}})\), and \(\text{v}\text{f}\text{y} = \text{v}\text{i}\text{y} + \text{g}\text{t}\) for final vertical velocity.

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.

initial velocity calculation
To start solving any projectile motion problem, it鈥檚 essential to first determine the initial velocity, denoted as \( v_i \). The initial velocity tells us how fast the projectile is moving when it begins its path and at what angle. In our exercise, the airplane is diving and releases the projectile at an angle of 53 degrees with the vertical. This gives the projectile an initial downward velocity.
trigonometric relations in physics
Understanding trigonometric relations is crucial in breaking down the components of motion. Whenever a velocity is given at an angle, trigonometry helps us find the horizontal and vertical components. Given the angle \( \theta = 53.0^\text{o} \), we use sine and cosine functions:

  • Vertical component: \( v_{iy} = v_i \sin(\theta) \)
  • Horizontal component: \( v_{ix} = v_i \cos(\theta) \)
The vertical component affects how long the projectile stays in the air, while the horizontal component determines the distance it travels.
horizontal and vertical components of velocity
Dividing the initial velocity into horizontal and vertical components makes solving projectile motion problems more manageable. The vertical component \( v_{iy} \) significantly affects the time the projectile is in motion due to gravity. The horizontal component \( v_{ix} \) remains constant throughout the flight since horizontal motion is not influenced by gravity.

In our problem:
  • For the vertical velocity component, we have: \( v_{iy} = v_i \sin(53.0^\text{o}) \). This will determine how quickly the projectile reaches the ground.
  • The horizontal velocity component is: \( v_{ix} = v_i \cos(53.0^\text{o}) \). This tells how far the projectile travels horizontally.
These components help to define the motion in both vertical and horizontal directions.
projectile motion equations
Projectile motion equations describe the motion of objects projected into the air and moving under the influence of gravity. For our problem, we use the following key equations:
  • The horizontal motion equation: \( x = v_{ix} t \). This describes the horizontal displacement.
  • The vertical motion equation: \( y_f = y_i + v_{iy} t - \frac{1}{2} g t^2 \). This tells us the vertical displacement.
  • Final vertical velocity: \( v_{fy} = v_{iy} + g t \)
These equations incorporate time (\( t \)), and the acceleration due to gravity (\( g \)), allowing us to predict the projectile鈥檚 path and calculate its final velocities and positions.
kinematics
Kinematics is the study of motion without considering the forces that cause it. It helps in describing the linear motion of objects, such as the projectile in our problem, through its parameters: displacement, velocity, and acceleration. In our scenario:

  • Initial conditions include the initial height (730 m) and time of flight (5.00 s).
  • The acceleration due to gravity (\( g \)) always acts downward at 9.8 m/s虏.
Through kinematic equations, we can solve for different aspects of the projectile motion, including how the projectile moves vertically and horizontally, its velocity components, and the overall travel time and distance.

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) What is the magnitude of the centripetal acceleration of an object on Earth's equator due to the rotation of Earth? (b) What would the period of rotation of Earth have to be for objects on the equator to have a centripetal acceleration with a magnitude of \(9.8 \mathrm{~m} / \mathrm{s}^{2} ?\)

In \(x y\) Plane The position \(\vec{r}\) of a particle moving in an \(x y\) plane is given by \(\vec{r}(t)=\left[\left(2.00 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3}-(5.00 \mathrm{~m} / \mathrm{s}) t\right] \hat{\mathrm{i}}+\) \(\left[(6.00 \mathrm{~m})-\left(7.00 \mathrm{~m} / \mathrm{s}^{4}\right) t^{4}\right] \hat{\mathrm{j}} .\) Calculate (a) \(\vec{r},(\mathrm{~b}) \vec{v}\), and \((\mathrm{c}) \vec{a}\) for \(t=2.00 \mathrm{~s}\)

Soccer Ball A soccer ball is Problem 17 . kicked from the ground with an initial speed of \(19.5 \mathrm{~m} / \mathrm{s}\) at an upward angle of \(45^{\circ} .\) A player \(55 \mathrm{~m}\) away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? Neglect air resistance. 19\. Stairway A ball rolls horizontally off the top of a stairway with a speed of \(1.52 \mathrm{~m} / \mathrm{s}\). The steps are \(20.3 \mathrm{~cm}\) high and \(20.3 \mathrm{~cm}\) wide. Which step does the ball hit first?

. Fast Bullets A rifle that shoots bullets at \(460 \mathrm{~m} / \mathrm{s}\) is to be aimed at a target \(45.7 \mathrm{~m}\) away and level with the rifle. How high above the target must the rifle barrel be pointed so that the bullet hits the target?

A train moving at a constant speed of \(60.0 \mathrm{~km} / \mathrm{h}\) moves east for \(40.0\) min. then in a direction \(50.0^{\circ}\) east of north for \(20.0 \mathrm{~min}\), and finally west for \(50.0 \mathrm{~min}\). What is the average velocity of the train during this trip?
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Translational Dynamics

Read Explanation

Modern Physics

Read Explanation

Astrophysics

Read Explanation

Electricity

Read Explanation

Magnetism and Electromagnetic Induction

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.