/*! 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 39 A slingshot fires a pebble from ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A slingshot fires a pebble from the top of a building at a speed of 14.0 m/s. The building is 31.0 m tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

Short Answer

Expert verified
(a) 28.3 m/s; (b) 24.6 m/s; (c) 28.3 m/s.

Step by step solution

01

Understanding the Problem

We need to find the final speed of a pebble fired at different initial angles from a building. We will use the principles of projectile motion and energy conservation.
02

Identifying Given Values

- Initial horizontal speed \( v_{0x} = 14.0 \text{ m/s} \).- Building height \( h = 31.0 \text{ m} \).- Gravitational acceleration \( g = 9.81 \text{ m/s}^2 \).
03

Using Energy Conservation - Horizontal Launch

For part (a), the pebble is launched horizontally. Its initial vertical speed \( v_{0y} = 0 \). The final speed \( v_f \) can be calculated by finding the vertical speed at impact due to gravity. Calculate the vertical speed: \[ v_y = \sqrt{2gh} = \sqrt{2\times9.81\times31.0} \approx 24.6 \text{ m/s} \].The final speed is \[ v_f = \sqrt{v_{0x}^2 + v_y^2} = \sqrt{14.0^2 + 24.6^2} \].
04

Solving for Speed - Horizontal Launch

Calculate the speed using:\[ v_f = \sqrt{14.0^2 + 24.6^2} = \sqrt{196 + 605.76} \approx \sqrt{801.76} \approx 28.3 \text{ m/s} \].
05

Using Energy Conservation - Vertical Up Launch

For part (b), the pebble is launched vertically upwards. The vertical position provides a maximum height before returning to the initial height, and then continuing to fall. Use energy conservation to find final vertical speed \( v_y = \sqrt{0 + 2gh} = 24.6 \text{ m/s} \), resulting in:\[ v_f = 24.6 \text{ m/s} \] since initial horizontal velocity is 0.
06

Using Energy Conservation - Vertical Down Launch

For part (c), the pebble is launched vertically downwards with the initial speed equal to 14.0 m/s. Calculate final speed:\[ v_f = \sqrt{v_{0x}^2 + 2gh} = \sqrt{14.0^2 + 2\times9.81\times31.0} \].
07

Solving for Speed - Vertical Down Launch

Calculate the speed using:\[ v_f = \sqrt{14.0^2 + 605.76} = \sqrt{196 + 605.76} \approx \sqrt{801.76} \approx 28.3 \text{ m/s} \].

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!

Key Concepts

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

Energy Conservation
Energy conservation is a fundamental principle that helps us understand projectile motion. It states that energy in a closed system remains constant. Specifically, the mechanical energy, which is the sum of potential and kinetic energy, is conserved in the absence of non-conservative forces like air resistance.

In the context of our problem, a pebble is fired from a building. Initially, it has kinetic energy due to its speed and potential energy due to its height. As it moves, energy shifts between potential and kinetic forms, but the total remains constant.
  • Kinetic Energy (KE): At any point, given by: \( KE = \frac{1}{2}mv^2 \)
  • Potential Energy (PE): Depending on height: \( PE = mgh \)
When the pebble hits the ground, all the potential energy has converted into kinetic energy, allowing us to calculate its final speed states.
Vertical Speed
Vertical speed is crucial in understanding the pebble's movement. It changes as the pebble travels due to gravity acting vertically. Initially, when the pebble is launched horizontally, the vertical speed is zero.

Over time, gravity increases the vertical speed, contributing to the final impact speed. For any free-falling object, we can calculate the vertical speed using:
  • Using acceleration due to gravity, vertical speed at impact: \( v_y = \sqrt{2gh} \)
  • This speed is independent of the object's horizontal motion.
Knowing the vertical speed helps determine the final speed when combined with horizontal components.
Gravitational Acceleration
Gravitational acceleration is a constant force acting on all objects near Earth's surface. Its standard value is \( g = 9.81 \text{ m/s}^2 \), directing downwards towards Earth's center.

In projectile motion, this constant acceleration influences the vertical motion of the object:
  • It affects the vertical speed over time, causing any vertically launched object to eventually fall back down.
  • Regardless of the initial horizontal speed, the vertical motion is universally affected by gravity.
This consistent gravitational pull allows us to predict how fast the pebble will move vertically upon impact and contributes to energy conservation calculations.
Horizontal Launch
A horizontal launch means that the initial vertical speed is zero. In this scenario, only the horizontal component of velocity is active at launch. This affects how we analyze projectile motion:

  • The horizontal speed remains constant because, in the absence of air resistance, no forces act horizontally.
  • Vertical speed starts at zero and increases due to gravitational acceleration.
During a horizontal launch, the final speed at which the pebble strikes the ground is a combination of unchanged horizontal speed and the vertical speed gained from gravity, computed using Pythagorean theorem.

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 sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled’s displacement, which is along the x axis. As a result, the kinetic energy of the sled increases by 38%. By what percentage would the sled’s kinetic energy have increased if this force had pointed \(62^{\circ}\) above the x axis?

A 75.0-kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant velocity from ground level to the floor above, a vertical height of 4.60 m. What is the work done on the man by (a) the gravitational force and (b) the escalator?

Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. What average power must the winch supply in order to accelerate a 184-kg ultralight glider from rest to 26.0 m/s over a horizontal distance of 48.0 m? Assume that friction and air resistance are negligible, and that the tension in the winch cable is constant.

A 55-kg box is being pushed a distance of 7.0 m across the floor by a force \(\overrightarrow{\mathbf{P}}\) whose magnitude is 160 \(\mathrm{N}\) . The force \(\overrightarrow{\mathbf{P}}\) is parallel to the displacement of the box. The coefficient of kinetic friction is 0.25.Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.

A pole-vaulter just clears the bar at 5.80 m and falls back to the ground. The change in the vaulter’s potential energy during the fall is \(-3.70 \times 10^{3} J .\) What is his weight?

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.