Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Playing shortstop, you pick up a ground ball and throw it to second base. The ball is thrown horizontally, with a speed of \(22 \mathrm{~m} / \mathrm{s}\), directly toward point \(A\) as shown in Figure 4.37. When the ball reaches the second baseman \(0.45 \mathrm{~s}\) later, it is caught at point B. (a) How far were you from the second baseman? (b) What is the distance of vertical drop, from \(A\) to \(B\) ?

Short Answer

Expert verified
(a) 9.9 m; (b) 0.99 m drop.

Step by step solution

01

Identify Given Information

We know the following information:- Horizontal velocity of the ball, \(v = 22 \text{ m/s}\)- Time for the ball to reach the second baseman, \(t = 0.45 \text{ s}\)
02

Calculate Horizontal Distance

The horizontal distance \(d\) can be calculated using the formula for constant speed: \[d = v \times t\]Substitute the known values:\(d = 22 \text{ m/s} \times 0.45 \text{ s} = 9.9 \text{ m}\)So, you were \(9.9\) meters away from the second baseman.
03

Determine Vertical Drop

To find the vertical drop, we need to use the formula for the distance fallen in free-fall, \(h = \frac{1}{2} g t^2\), where \(g\) is the acceleration due to gravity \(9.8 \text{ m/s}^2\).\[h = \frac{1}{2} \times 9.8 \text{ m/s}^2 \times (0.45 \text{ s})^2\]\(h \approx \frac{1}{2} \times 9.8 \times 0.2025 = 0.99 \text{ m}\)So, the ball vertically dropped approximately \(0.99\) meters from \(A\) to \(B\).

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 Vaia!

Key Concepts

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

Horizontal Velocity
Horizontal velocity refers to the constant speed at which an object moves along a horizontal path. It is a fundamental factor in projectile motion, where an object is launched into the air and follows a curved trajectory.
In the case of the baseball, the horizontal velocity is given as \(22 \text{ m/s}\), meaning that the ball travels at a constant speed of \(22 \text{ m/s}\) in the horizontal direction. This speed does not change as it moves towards the second baseman.
To find out how far the baseball travels, we use the formula for constant speed:
  • Distance \(d = v \times t\)
With \(v\) being the horizontal velocity and \(t\) the time taken. Substituting \(22 \text{ m/s}\) for the speed and \(0.45 \text{ s}\) for the time, the ball travels a distance of \(9.9 \text{ meters}\) horizontally. This calculation shows how horizontal velocity and time work together to determine distance.
Vertical Drop
The vertical drop in projectile motion refers to how much an object falls vertically due to gravity, while it moves horizontally. This drop is caused by the force of gravity acting on the object during its flight.
Even though the baseball is thrown horizontally, it simultaneously falls towards the ground. We calculate the vertical drop using the formula for free fall:
  • Vertical drop \(h = \frac{1}{2} g t^2\)
Here, \(g\) represents the acceleration due to gravity, which is \(9.8 \text{ m/s}^2\). By substituting \(0.45 \text{ s}\) for the time, the equation yields\[ h = \frac{1}{2} \times 9.8 \times (0.45)^2 \approx 0.99 \text{ m} \]This tells us that while the baseball travels horizontally, it drops approximately \(0.99 \text{ meters}\) vertically, demonstrating the simultaneous horizontal and vertical components of projectile motion.
Acceleration Due to Gravity
Acceleration due to gravity, often denoted by \(g\), is a key component in understanding projectile motion. It represents the rate at which an object accelerates towards the Earth's surface due to gravitational force.
On Earth, this value is roughly \(9.8 \text{ m/s}^2\). This means that in the absence of other forces, an object's velocity increases by \(9.8 \text{ m/s}\) every second it is in free fall.

Impact on the Baseball

In the exercise, gravity is what causes the baseball to drop \(0.99 \text{ meters}\) as it moves horizontally. Although the ball is thrown horizontally, it cannot overcome the pull of gravity, resulting in a vertical descent.
Gravity acts independently of any horizontal forces, causing the vertical component of motion to appear separate from the horizontal motion. Therefore, while the horizontal velocity remains constant, the vertical drop is dictated solely by gravity's pull on the ball, highlighting the dual nature of projectile trajectories.

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

Vector \(\overrightarrow{\mathbf{A}}\) points in the negative \(y\) direction and has a magnitude of \(5 \mathrm{~km}\). Vector \(\overrightarrow{\mathbf{B}}\) has a magnitude of \(15 \mathrm{~km}\) and points in the positive \(x\) direction. Use components to find the magnitude of (a) \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\), (b) \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\), and (c) \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}}\).

Describe How do you place the heads and tails of two vectors that you want to add?

A person on a cruise ship is doing laps on the promenade deck. On one portion of the track the person is moving north with a speed of \(3.8 \mathrm{~m} / \mathrm{s}\) relative to the ship. The ship moves east with a speed of \(12 \mathrm{~m} / \mathrm{s}\) relative to the water. What is the direction of motion of the person relative to the water?

As you hurry to catch your flight at the local airport, you encounter a moving walkway that is \(85 \mathrm{~m}\) long and has a speed of \(2.2 \mathrm{~m} / \mathrm{s}\) relative to the ground. If it takes you \(68 \mathrm{~s}\) to cover \(85 \mathrm{~m}\) when walking on the ground, how long will it take you to cover the same distance on the walkway? Assume that you walk with the same speed on the walkway as you do on the ground.

For each of the following quantities, indicate whether it is a scalar or a vector: (a) the time it takes you to run the \(100-m\) dash, (b) your displacement after running the \(100-\mathrm{m}\) dash, (c) your average velocity while running, (d) your average speed while running.

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.

Sign-up for free