JEE Main · 2014 · 06 Apr 2014 · Offline · Kinematics

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From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:

(A) $2 g H = n u^{2} (n - 2)$ ✓ Correct
(B) $g H = (n - 2) u^{2}$
(C) $2 g H = n^{2} u^{2}$
(D) $g H = (n - 2)^{2} u^{2}$

Answer: (A)

Solution

The time to reach the top is

t_{1} = \frac{u}{g}

, and the whole flight to the ground takes

t_{2} = n t_{1} = \frac{n u}{g}

. Measuring the downward displacement over the full flight,

- H = u t_{2} - \frac{1}{2} g t_{2}^{2} = \frac{n u ^{2}}{g} - \frac{n ^{2} u ^{2}}{2 g}

. Multiplying through by

2 g

and tidying,

2 g H = n u^{2} (n - 2)

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option

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