JEE Main 2020 · Physics · Kinematics and Laws of Motion · easy

A tower is 80 m high. A stone is dropped from its top and, at the same instant, a second stone is thrown vertically upward from the foot of the…

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A tower is

80\ \mathrm{m}

high. A stone is dropped from its top and, at the same instant, a second stone is thrown vertically upward from the foot of the tower. The two stones meet exactly at the mid-height of the tower. Taking

g = 10\ \mathrm{m\,s^{-2}}

, the speed of projection of the second stone is

A. $20\sqrt{2}\ \mathrm{m\,s^{-1}}$
B. $20\ \mathrm{m\,s^{-1}}$
C. $40\ \mathrm{m\,s^{-1}}$
D. $10\sqrt{2}\ \mathrm{m\,s^{-1}}$

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Answer: (A) $20\sqrt{2}\ \mathrm{m\,s^{-1}}$

The dropped stone falls

40\ \mathrm{m}

40 = \tfrac12 (10) t^2

gives

t = 2\sqrt2\ \mathrm{s}

. The thrown stone must rise

40\ \mathrm{m}

in the same time:

40 = u t - \tfrac12 g t^2 = u t - 40

, so

u t = 80

and

u = 80/(2\sqrt2) = 20\sqrt2\ \mathrm{m\,s^{-1}}

. Shortcut: the stones always meet when

t = H/u

(relative speed is

u

), so

H/2 = \tfrac12 g H^2/u^2

gives

u^2 = gH

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