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

A ball is shot from ground level over level ground at a certain initial speed. Figure 4-27 gives the range R of the ball versus its launch angle u0. Rank the three lettered points on the plot according to (a) the total flight time of the ball and (b) the ball’s speed at maximum height, greatest first.

Short Answer

Expert verified
  1. Rank:c>b>a
  1. Rank:a>b>c

Step by step solution

01

Given information

Curve of Range vs initial angle of projection

02

To understand the concept

It deals with the kinematic equations of motion in which the motion can be described with constant acceleration. Also, this problem is based on the concept of projectile motion. When a particle is thrown near the earth’s surface, it travels along a curved path under constant acceleration directed towards the center of the earth surface. This is a projectile motion problem, which contains two independent motions.

Horizontal motion with constant speed ax=0

Vertical motion with acceleration due to gravity a=9.8m/s2.

Formulae:

Range(R)=v0xTv0x=HorizontalcomponentofinitialspeedT-Timeofflight

03

(a) To rank the three lettered points on the plot according to the total time of flight of the ball

We know that,

R=V0xT=V0×cosθ0×T

Hence,

T=RV0cosθ0

Asθincreases, time of flight also increases, value ofcosθwould decrease. Therefore, rank according to time is:

Rank:c>b>a

04

(b) To rank the three lettered points on the plot according to the ball’s speed at maximum height, greatest first

At maximum height vertical component of velocity is zero, so speed is same as horizontal component of velocity which is same at any point.

Hence,

v=V0x=V0×cosθ0

Asθincreases, speed decreases. Therefore, rank according to speed at maximum height is:

Rank:a>b>c

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!

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 woman who can row a boat at 6.4 km/hin still water faces a long, straight river with a width of 6.4 km and a current of 3.2 km/h.i^ Letpoint directly across the river andj^point directly downstream. If she rows in a straight line to a point directly opposite her starting position, (a) at what angle toi^must she point the boat and (b) how long will she take? (c) How long will she take if, instead, she rowsdownthe river and then back to her starting point? (d) How long if she rows 3.2 kmupthe river and then back to her starting point? (e) At what angle toi^should she point the boat if she wants to cross the river in the shortest possible time? (f) How long is that shortest time?

In Fig. 4-54, a lump of wet putty moves in uniform circular motion as it rides at a radius of 20.0cmon the rim of a wheel rotating counterclockwise with a period of 5.00 ms .The lump then happens to fly off the rim at the 5 o’clock position (as if on a clock face). It leaves the rim at a height of h=1.20m from the floor and at a distance d=2.50 m from a wall. At what height on the wall does the lump hit?

A particle moves so that its position (in meters) as a function of time (in seconds) is r=i^+4t2j^+tk^. Write expressions for: (a) its velocity and (b) its acceleration as functions of time.

In 3.50h, a balloon drifts 21.5kmnorth, 9.70kmeast, and 2.88kmupward from its release point on the ground. Find (a) the magnitude of its average velocity. (b) the angle its average velocity makes with the horizontal.

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the spike is difficult. Suppose a ball is spiked from a height of2.30mwith an initial speed oflocalid="1654581187572" 20.0m/sat a downward angle oflocalid="1654581198810" 18.00°. How much farther on the opposite floor would it have landed if the downward angle were, instead,localid="1654581212383" 8.00°?

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