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Motion in A Plane question

2025 · 7 Apr · Shift 1 · Q62
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Motion in A Plane question

2025 · 7 Apr · Shift 1 · Q62

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
Two projectiles are fired from ground with same initial speeds from same point at angles (45∘+α)\left(45^{\circ}+\right.\alpha)(45∘+α) and (45∘−α)\left(45^{\circ}-\alpha\right)(45∘−α) with horizontal direction. The ratio of their times of flights is
  1. A
    1+tan⁡α1−tan⁡α\frac{1+\tan \alpha}{1-\tan \alpha}1−tanα1+tanα​
  2. B
    1+sin⁡2α1−sin⁡2α\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}1−sin2α1+sin2α​
  3. C
    1−tan⁡α1+tan⁡α\frac{1-\tan \alpha}{1+\tan \alpha}1+tanα1−tanα​
  4. D
    1
View written solutionFree

Correct answer: A

  1. Time of flight formula

For a projectile projected with speed uuu at angle θ\thetaθ from the horizontal, the time of flight is

T=2usin⁡θg.T=\frac{2u\sin\theta}{g}.T=g2usinθ​.

So here,

  • for the first projectile, θ1=45∘+α\theta_1=45^\circ+\alphaθ1​=45∘+α
  • for the second projectile, θ2=45∘−α\theta_2=45^\circ-\alphaθ2​=45∘−α

Hence,

T1=2usin⁡(45∘+α)g,T2=2usin⁡(45∘−α)g.T_1=\frac{2u\sin(45^\circ+\alpha)}{g}, \qquad T_2=\frac{2u\sin(45^\circ-\alpha)}{g}.T1​=g2usin(45∘+α)​,T2​=g2usin(45∘−α)​.

  1. Take the ratio

T1T2=sin⁡(45∘+α)sin⁡(45∘−α).\frac{T_1}{T_2}=\frac{\sin(45^\circ+\alpha)}{\sin(45^\circ-\alpha)}.T2​T1​​=sin(45∘−α)sin(45∘+α)​.

Now use

sin⁡(45∘+α)=12(cos⁡α+sin⁡α),\sin(45^\circ+\alpha)=\frac{1}{\sqrt{2}}(\cos\alpha+\sin\alpha),sin(45∘+α)=2​1​(cosα+sinα), sin⁡(45∘−α)=12(cos⁡α−sin⁡α).\sin(45^\circ-\alpha)=\frac{1}{\sqrt{2}}(\cos\alpha-\sin\alpha).sin(45∘−α)=2​1​(cosα−sinα).

Therefore,

T1T2=cos⁡α+sin⁡αcos⁡α−sin⁡α.\frac{T_1}{T_2}=\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}.T2​T1​​=cosα−sinαcosα+sinα​.

  1. Express in terms of tan⁡α\tan\alphatanα

Divide numerator and denominator by cos⁡α\cos\alphacosα:

T1T2=1+tan⁡α1−tan⁡α.\frac{T_1}{T_2}=\frac{1+\tan\alpha}{1-\tan\alpha}.T2​T1​​=1−tanα1+tanα​.

  1. Match with options

This matches Option A.

  1. Check other relevant option

Also note that

=\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}.$$ This is not generally the same expression as the ratio obtained directly in the standard form required here. The exact ratio from time of flight is $$\boxed{\frac{1+\tan\alpha}{1-\tan\alpha}}.$$
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