/*! 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 15 A ball is thrown straight up by ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A ball is thrown straight up by a passenger in a train that is moving with a constant velocity. Where would the ball land-back in his hands, in front of him, or behind him? Does your answer change if the train is accelerating in the forward direction? If yes, how?

Short Answer

Expert verified
Answer: (a) When the train is moving with constant velocity, the ball lands back in the passenger's hands. (b) When the train is accelerating in the forward direction, the ball lands behind the passenger.

Step by step solution

01

Case 1: Train moving with constant velocity

When the train is moving with a constant velocity, both the passenger and the ball share the same horizontal velocity. Since the ball is thrown straight up, it will experience only gravitational force acting on it vertically. No additional horizontal forces will be acting on the ball. As a result, the horizontal velocity of the ball remains constant throughout its ascent and descent. Therefore, the ball will continue to move horizontally in sync with the passenger, and the ball will land back in the passenger's hands.
02

Case 2: Train accelerating in the forward direction

When the train accelerates in the forward direction, the passenger and the ball will experience an additional force due to acceleration. At the moment when the ball is thrown straight up, the ball will have the same initial horizontal velocity as that of the passenger. However, the train continues to accelerate, while the ball does not experience any horizontal force after leaving the passenger's hand. Due to the absence of horizontal force acting on the ball, the ball's horizontal velocity will remain constant throughout its flight. In the meantime, the train and the passenger continue to accelerate. As they accelerate forward, the passenger gains horizontal velocity relative to the ball. Consequently, the ball will lag behind the passenger's moving position and will land behind the passenger.

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 91Ó°ÊÓ!

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 given initial speed of an ideal projectile, there is (are) \(\quad\) launch angle(s) for which the range of the projectile is the same. a) only one b) two different c) more than two but a finite number of d) only one if the angle is \(45^{\circ}\) but otherwise two different e) an infinite number of

A baseball is launched from the bat at an angle \(\theta_{0}=30^{\circ}\) with respect to the positive \(x\) -axis and with an initial speed of \(40 \mathrm{~m} / \mathrm{s}\), and it is caught at the same height from which it was hit. Assuming ideal projectile motion (positive \(y\) -axis upward), the velocity of the ball when it is caught is a) \((20.00 \hat{x}+34.64 \hat{y}) \mathrm{m} / \mathrm{s}\). b) \((-20.00 \hat{x}+34.64 \hat{y}) \mathrm{m} / \mathrm{s}\) c) \((34.64 \hat{x}-20.00 \hat{y}) \mathrm{m} / \mathrm{s}\) d) \((34.64 \hat{x}+20.00 \hat{y}) \mathrm{m} / \mathrm{s}\).

To attain maximum height for the trajectory of a projectile, what angle would you choose between \(0^{\circ}\) and \(90^{\circ}\), assuming that you can launch the projectile with the same initial speed independent of the launch angle. Explain your reasoning.

Salmon often jump upstream through waterfalls to reach their breeding grounds. One salmon came across a waterfall \(1.05 \mathrm{~m}\) in height, which she jumped in \(2.1 \mathrm{~s}\) at an angle of \(35^{\circ}\) to continue upstream. What was the initial speed of her jump?

A rock is thrown at an angle \(45^{\circ}\) below the horizontal from the top of a building. Immediately after release will its acceleration be greater than, equal to, or less than the acceleration due to gravity?

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.