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MCQEasyJEE 2023Projectile Motion

JEE Physics 2023 Question with Solution

The initial speed of a projectile fired from ground is uu. At the highest point during its motion, the speed of the projectile is 52u\frac{\sqrt{5}}{2}u. The time of flight of the projectile is:

  • A

    u2g\frac{u}{2g}

  • B

    ug\frac{u}{g}

  • C

    2ug\frac{2u}{g}

  • D

    5ug\frac{\sqrt{5}u}{g}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Initial speed is uu. At the highest point, the projectile speed is 52u\frac{\sqrt{5}}{2}u.

Find: The time of flight TT.

At the highest point, the vertical component of velocity is zero, so the speed there is equal to the horizontal component:

ucos⁡θ=52uu \cos \theta = \frac{\sqrt{5}}{2}u

Cancelling uu,

cos⁡θ=52\cos \theta = \frac{\sqrt{5}}{2}

Using sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1,

sin⁡θ=12\sin \theta = \frac{1}{2}

Hence,

θ=30∘\theta = 30^\circ

Now the time of flight of a projectile is

T=2usin⁡θgT = \frac{2u\sin\theta}{g}

Substituting sin⁡30∘=12\sin 30^\circ = \frac{1}{2},

T=2u⋅12g=ugT = \frac{2u \cdot \frac{1}{2}}{g} = \frac{u}{g}

Therefore, the time of flight is ug\frac{u}{g}. The correct option is B.

Common mistakes

  • Taking the speed at the highest point as zero is incorrect. Only the vertical component becomes zero there; the horizontal component remains unchanged. Use highest-point speed as ucos⁡θu\cos\theta.

  • Using the first solution line ucos⁡θ=32uu\cos\theta = \frac{\sqrt{3}}{2}u would be wrong for this question because the given highest-point speed is 52u\frac{\sqrt{5}}{2}u. Always match the equation with the stated data in the question.

  • Applying T=usin⁡θgT = \frac{u\sin\theta}{g} is incorrect because that gives only the time to reach the highest point. For total time of flight, use T=2usin⁡θgT = \frac{2u\sin\theta}{g}.

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