Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 49

A ball is thrown from a point \(1.0 \mathrm{m}\) above the ground. The initial velocity is \(19.6 \mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. (a) Find the maximum height of the ball above the ground. (b) Calculate the speed of the ball at the highest point in the trajectory.

Short Answer

Expert verified
Answer: The maximum height of the ball above the ground is 5.9 m, and its speed at the highest point of the trajectory is 16.968 m/s.

Step by step solution

01

Resolve the velocity vector into components

The initial velocity of the ball is given as \(19.6 \mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. To resolve the velocity vector into its horizontal (\(v_{x}\)) and vertical (\(v_{y}\)) components, we will use the trigonometric expressions for sine and cosine: \(v_x = v\cos(\theta)\) \(v_y = v\sin(\theta)\) \(v_x = 19.6\cos(30^{\circ}) = 16.968 \mathrm{m} / \mathrm{s}\) \(v_y = 19.6\sin(30^{\circ}) = 9.8 \mathrm{m} / \mathrm{s}\)
02

Calculate the time to the highest point

At the highest point of its trajectory, the vertical component of the ball's velocity will be zero. We can use the following equation of motion to find the time it takes for the ball to reach its highest point: \(v_y = u_y - gt\) Rearranging for \(t\): \(t = \frac{u_y - v_y}{g} = \frac{9.8 - 0}{9.81} = 1 \mathrm{s}\)
03

Calculate the maximum height above the ground

Now, we can use another equation of motion to find the highest point (\(h\)) of the ball above the ground: \(h = u_yt - \frac{1}{2}gt^2\) \(h = 9.8(1) - \frac{1}{2}(9.81)(1)^2 = 4.9 \mathrm{m}\) However, the ball is thrown from \(1.0 \mathrm{m}\) above the ground, so the maximum height of the ball above the ground is: \(H = 4.9 + 1 = 5.9 \mathrm{m}\)
04

Calculate the speed at the highest point

Since there is no air resistance, the horizontal component of the velocity remains constant throughout the motion. At the highest point of the trajectory, the vertical component of the velocity is zero, and the speed of the ball is equal to its horizontal component: \(v = v_x = 16.968 \mathrm{m} / \mathrm{s}\) Now, we have the answers to both parts of the problem: (a) The maximum height of the ball above the ground is \(5.9 \mathrm{m}\). (b) The speed of the ball at the highest point of the trajectory is \(16.968 \mathrm{m} / \mathrm{s}\).

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.

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

You have been employed by the local circus to plan their human cannonball performance. For this act, a spring-loaded cannon will shoot a human projectile, the Great Flyinski, across the big top to a net below. The net is located \(5.0 \mathrm{m}\) lower than the muzzle of the cannon from which the Great Flyinski is launched. The cannon will shoot the Great Flyinski at an angle of \(35.0^{\circ}\) above the horizontal and at a speed of $18.0 \mathrm{m} / \mathrm{s} .$ The ringmaster has asked that you decide how far from the cannon to place the net so that the Great Flyinski will land in the net and not be splattered on the floor, which would greatly disturb the audience. What do you tell the ringmaster? ( Wheractive: projectile motion)
A sailboat sails from Marblehead Harbor directly east for 45 nautical miles, then \(60^{\circ}\) south of east for 20.0 nautical miles, returns to an easterly heading for 30.0 nautical miles, and sails \(30^{\circ}\) east of north for 10.0 nautical miles, then west for 62 nautical miles. At that time the boat becomes becalmed and the auxiliary engine fails to start. The crew decides to notify the Coast Guard of their position. Using graph paper, ruler, and protractor, sketch a graphical addition of the displacement vectors and estimate their position.
A pilot starting from Athens, New York, wishes to fly to Sparta, New York, which is \(320 \mathrm{km}\) from Athens in the direction $20.0^{\circ} \mathrm{N}\( of \)\mathrm{E} .$ The pilot heads directly for Sparta and flies at an airspeed of \(160 \mathrm{km} / \mathrm{h}\). After flying for $2.0 \mathrm{h}$, the pilot expects to be at Sparta, but instead he finds himself \(20 \mathrm{km}\) due west of Sparta. He has forgotten to correct for the wind. (a) What is the velocity of the plane relative to the air? (b) Find the velocity (magnitude and direction) of the plane relative to the ground. (c) Find the wind speed and direction.
Imagine a trip where you drive along an east-west highway at $80.0 \mathrm{km} / \mathrm{h}$ for 45.0 min and then you turn onto a highway that runs \(38.0^{\circ}\) north of east and travel at \(60.0 \mathrm{km} / \mathrm{h}\) for 30.0 min. (a) What is your average velocity for the trip? (b) What is your average velocity on the return trip when you head the opposite way and drive \(38.0^{\circ}\) south of west at \(60.0 \mathrm{km} / \mathrm{h}\) for the first \(30.0 \mathrm{min}\) and then west at \(80.0 \mathrm{km} / \mathrm{h}\) for the last \(45.0 \mathrm{min} ?\)
Sheena can row a boat at \(3.00 \mathrm{mi} / \mathrm{h}\) in still water. She needs to cross a river that is 1.20 mi wide with a current flowing at $1.60 \mathrm{mi} / \mathrm{h} .$ Not having her calculator ready, she guesses that to go straight across, she should head \(60.0^{\circ}\) upstream. (a) What is her speed with respect to the starting point on the bank? (b) How long does it take her to cross the river? (c) How far upstream or downstream from her starting point will she reach the opposite bank? (d) In order to go straight across, what angle upstream should she have headed?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Quantum Physics

Read Explanation

Mechanics and Materials

Read Explanation

Circular Motion and Gravitation

Read Explanation

Famous Physicists

Read Explanation

Physics of Motion

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