/*! 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 96 A ball is thrown horizontally of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 96

A ball is thrown horizontally off the edge of a cliff with an initial speed of \(20.0 \mathrm{m} / \mathrm{s} .\) (a) How long does it take for the ball to fall to the ground 20.0 m below? (b) How long would it take for the ball to reach the ground if it were dropped from rest off the cliff edge? (c) How long would it take the ball to fall to the ground if it were thrown at an initial velocity of \(20.0 \mathrm{m} / \mathrm{s}\) but \(18^{\circ}\) below the horizontal?

Short Answer

Expert verified
Rearranging the equation and solving for \(t\) gives: \(t^2 = \dfrac{2(20)}{9.81}\) \(t^2 鈮� 4.08\) Therefore, \(t = \sqrt{4.08} 鈮� 2.02 \mathrm{s}\) In scenario (a) where the ball is thrown horizontally, it takes approximately 2.02 seconds for the ball to hit the ground.

Step by step solution

01

Scenario (a): Ball thrown horizontally

In this scenario, the ball is thrown horizontally off the edge of the cliff with an initial speed of 20.0 m/s. Since the motion of the ball is happening both horizontally and vertically, we'll have separate equations for horizontal and vertical planes. First, we need to find the time it takes for the ball to fall 20.0 m below. Using the vertical motion equation with an initial velocity of 0 (due to no vertical motion initially) and with acceleration due to gravity \(\mathrm{g} = 9.81 \mathrm{m} / \mathrm{s}^2\), we have: \(y = v_{0y} t - 0.5 gt^2\) Plugging in the values, we get: \(20 = 0 - 0.5(9.81)t^2\) Now, we need to solve for \(t\).

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!

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 ball is thrown from a point \(1.0 \mathrm{m}\) above the ground. The initial velocity is \(19.6 \mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. (a) Find the maximum height of the ball above the ground. (b) Calculate the speed of the ball at the highest point in the trajectory.
Orville walks 320 m due east. He then continues walking along a straight line, but in a different direction, and stops \(200 \mathrm{m}\) northeast of his starting point. How far did he walk during the second portion of the trip and in what direction?
Two displacement vectors each have magnitude \(20 \mathrm{km}\) One is directed \(60^{\circ}\) above the \(+x\) -axis; the other is directed \(60^{\circ}\) below the \(+x\) -axis. What is the vector sum of these two displacements? Use graph paper to find your answer.
A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is \(40 \mathrm{m} / \mathrm{s}\) at an angle of \(37^{\circ}\) with respect to the horizontal and the cannonball is $7.0 \mathrm{m}$ above the ground at this time. (a) What is the maximum height above the ground reached by the cannonball? (b) Assuming the cannonball makes it over the castle walls and lands back down on the ground, at what horizontal distance from its release point will it land? (c) What are the \(x\) - and y-components of the cannonball's velocity just before it lands? The \(y\) -axis points up.
A tennis ball is thrown horizontally from an elevation of \(14.0 \mathrm{m}\) above the ground with a speed of \(20.0 \mathrm{m} / \mathrm{s}\) (a) Where is the ball after 1.60 s? (b) If the ball is still in the air, how long before it hits the ground and where will it be with respect to the starting point once it lands?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Force

Read Explanation

Linear Momentum

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Energy Physics

Read Explanation

Electricity

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.