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MCQEasyJEE 2025Projectile Motion

JEE Physics 2025 Question with Solution

Two balls with the same mass and initial velocity are projected at different angles in such a way that the maximum height reached by the first ball is 88 times higher than that of the second ball. T1T_1 and T2T_2 are the total flying times of the first and second ball, respectively, then the ratio of T1T_1 and T2T_2 is:

  • A

    2:12 : 1

  • B

    2:1\sqrt{2} : 1

  • C

    4:14 : 1

  • D

    22:12\sqrt{2} : 1

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Two projectiles have the same initial velocity. The maximum height of the first ball is 88 times that of the second ball.

Find: The ratio T1T2\frac{T_1}{T_2} of their total times of flight.

For projectile motion,

H=vi2sin⁡2θ2gH = \frac{v_i^2 \sin^2 \theta}{2g}

and

T=2visin⁡θgT = \frac{2v_i \sin \theta}{g}

Using the given condition,

H1H2=8\frac{H_1}{H_2} = 8

So,

vi2sin⁡2θ12gvi2sin⁡2θ22g=8\frac{\frac{v_i^2 \sin^2 \theta_1}{2g}}{\frac{v_i^2 \sin^2 \theta_2}{2g}} = 8

Hence,

sin⁡2θ1sin⁡2θ2=8\frac{\sin^2 \theta_1}{\sin^2 \theta_2} = 8

Taking square root,

sin⁡θ1sin⁡θ2=8=22\frac{\sin \theta_1}{\sin \theta_2} = \sqrt{8} = 2\sqrt{2}

Now,

T1T2=2visin⁡θ1g2visin⁡θ2g=sin⁡θ1sin⁡θ2=22\frac{T_1}{T_2} = \frac{\frac{2v_i \sin \theta_1}{g}}{\frac{2v_i \sin \theta_2}{g}} = \frac{\sin \theta_1}{\sin \theta_2} = 2\sqrt{2}

Therefore, the ratio of the total flying times is 22:12\sqrt{2} : 1. The correct option is D.

Direct Relation Between Height and Time of Flight

Given: Same initial velocity for both balls and H1=8H2H_1 = 8H_2.

Find: The ratio T1:T2T_1 : T_2.

Since

H∝sin⁡2θH \propto \sin^2 \theta

and

T∝sin⁡θT \propto \sin \theta

for the same initial velocity, it follows that

T∝HT \propto \sqrt{H}

Therefore,

T1T2=H1H2=8=22\frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}} = \sqrt{8} = 2\sqrt{2}

Hence, the required ratio is 22:12\sqrt{2} : 1. The correct option is D.

Common mistakes

  • Using H∝sin⁡θH \propto \sin \theta instead of H∝sin⁡2θH \propto \sin^2 \theta is incorrect because maximum height depends on the square of the vertical component of velocity. Use H=vi2sin⁡2θ2gH = \frac{v_i^2 \sin^2 \theta}{2g}.

  • Comparing T1T_1 and T2T_2 directly from the height ratio as 8:18:1 is wrong because time of flight depends on sin⁡θ\sin \theta, not on sin⁡2θ\sin^2 \theta. Take the square root of the height ratio before relating the times.

  • Forgetting that both balls have the same initial velocity can lead to retaining viv_i in the ratio. Since the initial velocity is the same, it cancels out when forming both H1H2\frac{H_1}{H_2} and T1T2\frac{T_1}{T_2}.

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