JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Kinematics

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in

t_{1}

. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in

t_{2}

. Time required to reach the ground, if it is dropped from the top of the tower, is :

(A) $t_{1} t_{2}$ ✓ Correct
(B) $\frac{t _{1}}{t _{2}}$
(C) $t_{1} - t_{2}$
(D) $t_{1} + t_{2}$

Answer: (A)

Solution

Let the tower height be

h

and the launch speed

u

. For the upward throw reaching the ground in

t_{1}

and the downward throw in

t_{2}

, solving the respective equations of motion for

u

and

h

gives

t_{1} t_{2} = \frac{2 h}{g}

. Dropping the body from rest,

h = \frac{1}{2} g t^{2}

, so

t = 2 h / g = t_{1} t_{2}

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option

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