Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 50

A projectile is launched at a \(60^{\circ}\) angle above the horizontal on level ground. The change in its velocity between launch and just before landing is found to be \(\Delta \vec{v}=\vec{v}_{\text {landing }}-\vec{v}_{\text {launch }}=-20 \hat{y} \mathrm{~m} / \mathrm{s}\). What is the initial velocity of the projectile? What is its final velocity just before landing?

Short Answer

Expert verified
Answer: The final velocity of the projectile just before landing is given by \(\vec{v}_{landing} = \left( v_{0}\cos(60^{\circ})\right)\hat{x} + \left(v_{0}\sin(60^{\circ}) -20 \,\text{m/s}\right)\hat{y}\), where \(v_{0}\) is the initial velocity of the projectile.

Step by step solution

01

Determine the vertical and horizontal components of the change in velocity

The change in velocity is given as \(\Delta \vec{v} = -20\hat{y}\,\text{m/s}\). Since the projectile is launched on level ground and lands on level ground, there is no horizontal change in velocity, so \(\Delta v_x = 0\). The vertical change in velocity is given as \(\Delta v_y = -20\,\text{m/s}\).
02

Write the projectile motion equations for vertical and horizontal components

We can write the equations for the vertical and horizontal motion components. Horizontal motion: \(v_{x} = v_{0x}\) Vertical motion: \(v_{y} = v_{0y} - gt\) where \(v_{0y}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (\(9.8\,\text{m/s}^2\)) and \(t\) is the time of flight.
03

Find the initial vertical velocity component from the change in vertical velocity

Using the equation for vertical motion, we can relate the change in vertical velocity to the initial velocity, \(v_y - v_{0y} = -gt\) Since we know the change in vertical velocity, we can write: \(v_{0y} - (-20\,\text{m/s}) = -gt\) \(v_{0y} = -20\,\text{m/s} + gt\)
04

Use the launch angle to find the initial horizontal velocity component

We know the projectile is launched at a \(60^{\circ}\) angle above the horizontal, so we can find the initial horizontal velocity component as: \(v_{0x} = v_{0} \cos(60^{\circ})\) Where \(v_{0}\) is the initial velocity of the projectile.
05

Use the launch angle to find the initial vertical velocity component

Similarly, we can find the initial vertical velocity component using the same angle: \(v_{0y} = v_{0} \sin(60^{\circ})\)
06

Equating initial vertical velocity components to find the time of flight

From step 3 and step 5, we have, \(v_{0} \sin(60^{\circ}) = -20\,\text{m/s} + gt\) Solve for \(t\): \(t = \frac{v_{0} \sin(60^{\circ}) +20\,\text{m/s}}{g}\)
07

Find the initial velocity of the projectile

Now, we can square both horizontal and vertical components of initial velocity and sum them to find the initial velocity: \(v_{0}^2 = v_{0x}^2 + v_{0y}^2 \) \(v_{0}^2 = \left(v_{0}\cos(60^{\circ})\right)^2 + \left(v_{0}\sin(60^{\circ})\right)^2\) Solve for \(v_{0}\): \(v_{0} = \sqrt{\left(v_{0}\cos(60^{\circ})\right)^2 + \left(v_{0}\sin(60^{\circ})\right)^2}\)
08

Calculate the final velocity just before landing

To calculate the final velocity just before landing, we account for the fact that the horizontal component remains unchanged throughout the flight and the vertical component has changed as given: \(\vec{v}_{landing} = \left( v_{0}\cos(60^{\circ})\right)\hat{x} + \left(v_{0}\sin(60^{\circ}) -20 \,\text{m/s}\right)\hat{y}\)

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

In ideal projectile motion, the velocity and acceleration of the projectile at its maximum height are, respectively, a) horizontal, vertical c) zero, zero. downward. d) zero, vertical downward. b) horizontal, zero. e) zero, horizontal.

A projectile is launched from the top of a building with an initial velocity of \(30 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) above the horizontal. The magnitude of its velocity at \(t=5 \mathrm{~s}\) after the launch is a) \(-23.0 \mathrm{~m} / \mathrm{s}\) c) \(15.0 \mathrm{~m} / \mathrm{s}\) e) \(50.4 \mathrm{~m} / \mathrm{s}\) b) \(7.3 \mathrm{~m} / \mathrm{s}\) d) \(27.5 \mathrm{~m} / \mathrm{s}\)

10.0 seconds after being fired, a cannonball strikes a point \(500 . \mathrm{m}\) horizontally from and \(100 . \mathrm{m}\) vertically above the point of launch. a) With what initial velocity was the cannonball launched? b) What maximum height was attained by the ball? c) What is the magnitude and direction of the ball's velocity just before it strikes the given point?

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 initial 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?

During a jaunt on your sailboat, you sail \(2.00 \mathrm{~km}\) east, \(4.00 \mathrm{~km}\) southeast, and 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

Electricity

Read Explanation

Space Physics

Read Explanation

Medical Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Nuclear Physics

Read Explanation

Dynamics

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