JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Kinematics

The initial speed of a projectile fired from ground is

u

. At the highest point during its motion, the speed of projectile is

\frac{3}{2} u

. The time of flight of the projectile is:

(A) $\frac{u}{2 g}$
(B) $\frac{u}{g}$ ✓ Correct
(C) $\frac{2 u}{g}$
(D) $\frac{3 u}{g}$

Answer: (B)

Solution

At the top of the trajectory the vertical velocity is zero, so the speed there is just the horizontal component

u cos θ

. Equating this to the given top speed,

u cos θ = \frac{3}{2} u

, gives

cos θ = \frac{3}{2}

, so the projection angle is

θ = 3 0^{\circ}

. The time of flight is governed by the vertical motion,

T = \frac{2 u sin θ}{g} = \frac{2 u sin 3 0 ^{\circ}}{g} = \frac{2 u \cdot \frac{1}{2}}{g} = \frac{u}{g}

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option

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