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MCQEasyJEE 2025Projectile Motion

JEE Physics 2025 Question with Solution

Two projectiles are fired with the same initial speed from the same point on the ground at angles of (45∘−α)\left(45^\circ - \alpha\right) and (45∘+α)\left(45^\circ + \alpha\right), respectively, with the horizontal direction. The ratio of their maximum heights attained is:

  • A

    1−tan⁡α1+tan⁡α\frac{1 - \tan \alpha}{1 + \tan \alpha}

  • B

    1−sin⁡2α1+sin⁡2α\frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}

  • C

    1+sin⁡2α1−sin⁡2α\frac{1 + \sin 2\alpha}{1 - \sin 2\alpha}

  • D

    1+sin⁡α1−sin⁡α\frac{1 + \sin \alpha}{1 - \sin \alpha}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two projectiles are fired with the same initial speed at angles (45∘−α)\left(45^\circ - \alpha\right) and (45∘+α)\left(45^\circ + \alpha\right).

Find: The ratio of their maximum heights.

For projectile motion, the maximum height is

H=v02sin⁡2θ2gH = \frac{v_0^2 \sin^2 \theta}{2g}

So, for the two projectiles,

H1=v02sin⁡2(45∘−α)2gH_1 = \frac{v_0^2 \sin^2 \left(45^\circ - \alpha\right)}{2g}

and

H2=v02sin⁡2(45∘+α)2gH_2 = \frac{v_0^2 \sin^2 \left(45^\circ + \alpha\right)}{2g}

Therefore,

H1H2=sin⁡2(45∘−α)sin⁡2(45∘+α)\frac{H_1}{H_2} = \frac{\sin^2 \left(45^\circ - \alpha\right)}{\sin^2 \left(45^\circ + \alpha\right)}

Using

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B

we get

sin⁡(45∘−α)=22(cos⁡α−sin⁡α)\sin\left(45^\circ - \alpha\right) = \frac{\sqrt{2}}{2}(\cos \alpha - \sin \alpha)

and

sin⁡(45∘+α)=22(cos⁡α+sin⁡α)\sin\left(45^\circ + \alpha\right) = \frac{\sqrt{2}}{2}(\cos \alpha + \sin \alpha)

Thus,

H1H2=(cos⁡α−sin⁡αcos⁡α+sin⁡α)2\frac{H_1}{H_2} = \left(\frac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}\right)^2

Now,

(cos⁡α−sin⁡α)2=cos⁡2α+sin⁡2α−2sin⁡αcos⁡α=1−sin⁡2α(\cos \alpha - \sin \alpha)^2 = \cos^2 \alpha + \sin^2 \alpha - 2\sin \alpha \cos \alpha = 1 - \sin 2\alpha

Similarly,

(cos⁡α+sin⁡α)2=cos⁡2α+sin⁡2α+2sin⁡αcos⁡α=1+sin⁡2α(\cos \alpha + \sin \alpha)^2 = \cos^2 \alpha + \sin^2 \alpha + 2\sin \alpha \cos \alpha = 1 + \sin 2\alpha

Hence,

H1H2=1−sin⁡2α1+sin⁡2α\frac{H_1}{H_2} = \frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}

Therefore, the ratio of the maximum heights is 1−sin⁡2α1+sin⁡2α\frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}. The correct option is B.

Common mistakes

  • Using the range formula instead of the maximum height formula is incorrect because the question asks about the highest point reached, not horizontal distance. Use H=v02sin⁡2θ2gH = \frac{v_0^2 \sin^2 \theta}{2g}.

  • Forgetting to square the sine term is a common mistake. The maximum height depends on sin⁡2θ\sin^2 \theta, not on sin⁡θ\sin \theta alone.

  • Applying the identity for sin⁡(45∘±α)\sin(45^\circ \pm \alpha) incorrectly leads to sign errors. Write both expansions carefully before forming the ratio.

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