For angles of projection of projectile at angle (45^ -θ) and (45^ +θ) , the horizontal range described by the projectile are in the ratio of:

Correct answer: B. 1: 1 Horizontal range =R= (u^2 ^2 θ)/(g) When angle of projection is (45^ ^ -θ) then aligned R_1 & = u^2 2(45^ ^ -θ) g & = u^2 (90^ ^ -θ) g & =(u^2 2 θ)/(g) aligned Now (R_1)/(R_2)=(u^2 (2 θ)/(8))/(u^2 (2 θ)/(g))=1: 1

NEET 2006

Exam: NEET
Year: 2006
Subject: Physics
Chapter: Motion in Two Dimensions

NEET 2006 · Physics · Motion in Two Dimensions

For angles of projection of projectile at angle (45^ -θ) and (45^...

For angles of projection of projectile at angle

\left(45^{\circ}-\theta\right) and \left(45^{\circ}+\theta\right)

, the horizontal range described by the projectile are in the ratio of:

Options

A $2: 1$
B $1: 1$
C2:3
D $1: 2$

Correct answer

B. $1: 1$

Step-by-step solution

Horizontal range =R=

\frac{u^2 \sin ^2 \theta}{g}

When angle of projection is

\left(45^{{}^{\circ}}-\theta\right)

then

\begin{aligned}R_1 & =\frac{u^2 \sin 2\left(45^{{}^{\circ}}-\theta\right)}{g}\\ & =\frac{u^2 \sin \left(90^{{}^{\circ}}-\theta\right)}{g}\\ & =\frac{u^2 \cos 2 \theta}{g} \end{aligned}

Now

\quad \frac{R_1}{R_2}=\frac{u^2 \cos \frac{2 \theta}{8}}{u^2 \cos \frac{2 \theta}{g}}=1: 1

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