Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 4

An outfielder throws the baseball to first base, located \(80 \mathrm{~m}\) away from the fielder, with a velocity of \(45 \mathrm{~m} / \mathrm{s}\). At what launch angle above the horizontal should he throw the ball for the first baseman to catch the ball in \(2 \mathrm{~s}\) at the same height? a) \(50.74^{\circ}\) c) \(22.7^{\circ}\) e) \(12.6^{\circ}\) b) \(25.4^{\circ}\) d) \(18.5^{\circ}\)

Short Answer

Expert verified
The initial velocity of the thrown ball is 45 m/s. Answer: The launch angle for the outfielder must be approximately \(12.63^{\circ}\).

Step by step solution

01

Write down the given information

We are given: - The horizontal distance (range) to the first baseman: \(R = 80\, \mathrm{m}\) - The time taken to catch the ball: \(t = 2\, \mathrm{s}\) - The initial velocity of the thrown ball: \(v_0 = 45\, \mathrm{m/s}\) - The launch angle: \(\theta\) <-- We need to find this.
02

Determine the horizontal and vertical components of the initial velocity

The initial velocity can be split into horizontal and vertical components. Horizontal component: \(v_{0x} = v_0 \cos(\theta)\) Vertical component: \(v_{0y} = v_0 \sin(\theta)\)
03

Writing down the equations for horizontal and vertical motion

Horizontal motion: Since horizontal motion has constant velocity, we can use the equation \(R = v_{0x}t\). Vertical motion: Since the ball is caught at the same height, the vertical displacement is zero. The acceleration due to gravity (\(g\)) is acting downward. We can use the equation \(v = v_{0y} - gt\), where \(v\) is the final vertical velocity of the baseball.
04

Use horizontal motion equation to find the launch angle

We will first use the horizontal motion equation to find the cosine value of the launch angle: \(R = v_{0x}t = v_0 \cos(\theta)*t\) We can now find the cosine value of the launch angle: \(\cos(\theta) = \frac{R}{v_0 * t}\) Plug in the known values: \(\cos(\theta) = \frac{80}{45*2} = 0.8889\) Now, find the launch angle: \(\theta = \arccos(0.8889) \approx 27.57^{\circ}\)
05

Use vertical motion equation to find the final vertical velocity

We'll now use the vertical motion equation to find the final vertical velocity: \(v = v_{0y} - gt = v_0 \sin(\theta) - gt\) Since the ball is caught at the same height as it was thrown, the final vertical velocity \(v\) is equal in magnitude but opposite in direction to the initial vertical velocity, \(v_{0y}\): \(v = -v_0 \sin(\theta)\)
06

Find the sine value of the launch angle

We will now use the relation derived in step 5 to find the sine value of the launch angle. \(-v_0 \sin(\theta) = v_0 \sin(\theta) - gt\) \(\sin(\theta) = \frac{gt}{2v_0}\) Plug in the known values (use \(g = 9.81\, \mathrm{m/s}^2\)): \(\sin(\theta) = \frac{2*9.81}{2*45} = 0.2180\) Now, find the launch angle: \(\theta = \arcsin(0.2180) \approx 12.63^{\circ}\)
07

Choose the correct launch angle

Comparing the two launch angles calculated in Steps 4 and 6, we choose the lesser angle as the solution, since the greater angle would result in the ball reaching a greater height and taking more than 2 seconds to reach the first baseman. Thus, the correct launch angle is \(\boxed{12.63^{\circ}}\), which corresponds to option (e).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

In a projectile motion, the horizontal range and the maximum height attained by the projectile are equal. a) What is the launch angle? b) If everything else stays the same, how should the launch angle, \(\theta_{0},\) of a projectile be changed for the range of the projectile to be halved?

A projectile is launched at an angle of \(45^{\circ}\) above the horizontal. What is the ratio of its horizontal range to its maximum height? How does the answer change if the initial speed of the projectile is doubled?

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?

What is the magnitude of an object's average velocity if an object moves from a point with coordinates \(x=2.0 \mathrm{~m}\) \(y=-3.0 \mathrm{~m}\) to a point with coordinates \(x=5.0 \mathrm{~m}, y=-9.0 \mathrm{~m}\) in a time interval of \(2.4 \mathrm{~s} ?\)

Rain is falling vertically at a constant speed of \(7.00 \mathrm{~m} / \mathrm{s}\). At what angle from the vertical do the raindrops appear to be falling to the driver of a car traveling on a straight road with a speed of \(60.0 \mathrm{~km} / \mathrm{h} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Nuclear Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Oscillations

Read Explanation

Fundamentals of Physics

Read Explanation

Measurements

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.

Sign-up for free