/*! 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 40 A golf ball is released from res... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A golf ball is released from rest from the top of a very tall building. Neglecting air resistance, calculate (a) the position and (b) the velocity of the ball after \(1.00,2.00,\) and \(3.00 \mathrm{s}\).

Short Answer

Expert verified
The ball falls the following distances: about 4.9 m, 19.6 m, and 44.1 m, and it's moving at the speeds: 9.8 m/s, 19.6 m/s, and 29.4 m/s after 1.00, 2.00, and 3.00 seconds, respectively.

Step by step solution

01

Identify Variables

Here are the variables in the problem: initial velocity (u) = 0 (since the ball is released from rest), time (t) = 1.00, 2.00, and 3.00 seconds, acceleration due to gravity (a) = \(-9.8 \mathrm{m/s^2}\) which is always constant and acts in the downward direction, final velocity (v), and distance (d).
02

Calculate Distance (position) at t = 1, 2, 3 seconds

Use the equation of motion that relates distance, initial velocity, time, and acceleration: \(d = ut + \frac{1}{2}at^2\). Since u = 0, the first term cancels out which gives: \(d = \frac{1}{2}at^2\). Plug in the times t = 1.00 s, 2.00 s, 3.00 s and a= \(-9.8 \mathrm{m/s^2}\) to get the distances corresponding to these times.
03

Calculate Velocity at t = 1, 2, 3 seconds

Use the equation of motion that relates velocity, initial velocity, acceleration, and time: \(v = u + at\). Here, u = 0, so v = at. Plug in the times t = 1.00, 2.00, 3.00 seconds and a= \(-9.8 \mathrm{m/s^2}\) to get the velocities corresponding to these times.

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.

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It's all about describing how things move—think position, velocity, and acceleration. When we talk about a golf ball being dropped from a building, as in the provided exercise, kinematics gives us the tools to predict where and how fast the ball will be at any given moment.

To do this, we use a series of kinematic equations that relate these quantities to each other. In the case of the golf ball, we are initially concerned with its position over time. Since the ball starts from rest, its initial velocity (commonly noted as 'u') is zero. Then, to figure out its position after 1, 2, and 3 seconds, we'd use the kinematic equation for displacement under constant acceleration, which, due to gravity, is present in our scenario.
Equations of Motion
The equations of motion are a set of formulas that provide a simple way of predicting the position and velocity of an object moving under constant acceleration. When an object, such as our golf ball, is dropped from a height, it accelerates downwards—thanks to gravity—at a constant rate. This allows us to apply these equations to predict the ball's future location and velocity.

Calculating Position

One such equation is used for computing the distance (or position) that the ball has fallen: \(d = ut + \frac{1}{2}at^2\). In this formula, 'd' represents distance, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time. Since the ball starts from rest ('u' being 0), the equation simplifies to just \(d = \frac{1}{2}at^2\). By plugging in the values, we'd get the ball's position at various times.

Calculating Velocity

Similarly, another equation gives us the final velocity ('v') after some time: \(v = u + at\). With the initial velocity being zero, the equation simplifies, and we can calculate how fast the ball is moving after 1, 2, and 3 seconds.

These equations are powerful as they allow us to make accurate predictions about an object's motion without knowing the forces involved—an essential skill in physics problem-solving.
Free Fall
Free fall is a specific type of motion that occurs when an object is only influenced by gravity. The key point about free fall is that it doesn't matter what the object is; everything falls at the same rate when air resistance is negligible. This is precisely the situation with our golf ball.

In this scenario, the golf ball undergoes free fall from the top of the building, and its acceleration 'a' is due to the force of gravity alone. Gravity pulls it downward at a constant acceleration of approximately \(-9.8 \mathrm{m/s^2}\), which is the value used in our equations of motion. This negative sign indicates the direction of the acceleration—towards the center of the Earth.

Understanding free fall is crucial because it not only simplifies the calculations by ensuring a constant acceleration but also provides a foundational concept that relates to other more complex physics scenarios. The idea that all objects in free fall (ignoring air resistance) experience the same gravitational pull regardless of their mass can sometimes seem unintuitive, but it's a cornerstone of classical physics.

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

Kathy Kool buys a sports car that can accelerate at the rate of \(4.90 \mathrm{m} / \mathrm{s}^{2} .\) She decides to test the car by racing with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 1.00 s before Kathy. If Stan moves with a constant acceleration of \(3.50 \mathrm{m} / \mathrm{s}^{2}\) and Kathy maintains an acceleration of \(4.90 \mathrm{m} / \mathrm{s}^{2},\) find. (a) the time at which Kathy overtakes Stan, (b) the distance she travels before she catches him, and (c) the speeds of both cars at the instant she overtakes him.

A ball is thrown directly downward, with an initial speed of \(8.00 \mathrm{m} / \mathrm{s},\) from a height of \(30.0 \mathrm{m} .\) After what time interval does the ball strike the ground?

A 745 i BMW car can brake to a stop in a distance of \(121 \mathrm{ft}\). from a speed of \(60.0 \mathrm{mi} / \mathrm{h} .\) To brake to a stop from a speed of \(80.0 \mathrm{mi} / \mathrm{h}\) requires a stopping distance of \(211 \mathrm{ft}\) What is the average braking acceleration for (a) \(60 \mathrm{mi} / \mathrm{h}\) to rest, (b) \(80 \mathrm{mi} / \mathrm{h}\) to rest, \((\mathrm{c}) 80 \mathrm{mi} / \mathrm{h}\) to \(60 \mathrm{mi} / \mathrm{h} ?\) Express the answers in \(\mathrm{mi} / \mathrm{h} / \mathrm{s}\) and in \(\mathrm{m} / \mathrm{s}^{2}\) .

A motorist drives along a straight road at a constant speed of \(15.0 \mathrm{m} / \mathrm{s} .\) Just as she passes a parked motorcycle police officer, the officer starts to accelerate at \(2.00 \mathrm{m} / \mathrm{s}^{2}\) to overtake her. Assuming the officer maintains this acceleration, (a) determine the time it takes the police officer to reach the motorist. Find (b) the speed and (c) the total displacement of the officer as he overtakes the motorist.

A car has an initial velocity \(v_{0}\) when the driver sees an obstacle in the road in front of him. His reaction time is \(\Delta t_{r}\) and the braking acceleration of the car is \(a\). Show that the total stopping distance is $$s_{\text {stop }}=v_{0} \Delta t_{r}-v_{0}^{2} / 2 a$$,Remember that \(a\) is a negative number.

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.