/*! 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} Q. 33 Show that a ball thrown with an ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that a ball thrown with an angle of elevation of 45∘ will travel farther than if it is thrown at any other angle.

Short Answer

Expert verified

The function rt=xt,ytrt=v0cosθt,-12gt2+v0sinθt models the motion of the ball. The ball will hit the ground when the y-component is zero. This will occur when t=2v0sinθg. At this time it will have travelled v0sinθcosθg units horizontally. Thus, x-component will be maximized whenθ=45∘.

Step by step solution

01

Step 1. Given information,

Consider the given question,

Angle of elevation is45∘.

02

Step 2. Find the distance traveled by the ball.

Assume the required function rt=xt,yt.

Assume rt=v0cosθt,-12gt2+v0sinθtmodels the motion of the ball.

If the ball hit the ground, y-component is zero.

-12gt2+v0sinθt=0v0sinθ=12gtt=2v0sinθg

The ball will travellocalid="1649613493436" v0sinθcosθgunits horizontally.

03

Step 3. Find when the x-component is maximum.

As v0,gare fixed, we fixed the value of θ at which the x-component is maximum.

Assume fθ=sinθcosθ. Then,

role="math" localid="1649613343908" fθ=122sinθcosθfθ=12sin2θf'0=122cos2θf'0=cos2θ

Let f'θ=0. Then,

cos2θ=02θ=90∘θ=45∘

Therefore, v0sinθcosθ9will be maximum when θ=45∘.

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!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.