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:

Motion in A Straight Line PYQ — JEE Main 2014 · Shift 0 · Q73 with Solution

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Correct answer: C

When the particle is thrown upward with speed $u$ , the time taken to reach the top is

$t_1=\frac{u}{g}$

because at the highest point, final velocity becomes zero.

Given that the total time taken to hit the ground is $n$ times the time to reach the highest point,

$T=n t_1=n\frac{u}{g}$

Take upward as positive.

Initial position: top of tower

Final position: ground, which is at displacement $-H$ from the top.

Using

$s=ut-\frac{1}{2}gt^2$

we get

$-H=uT-\frac{1}{2}gT^2$

Substitute $T=n\frac{u}{g}$ :

$-H=u\left(n\frac{u}{g}\right)-\frac{1}{2}g\left(n\frac{u}{g}\right)^2$

$-H=\frac{nu^2}{g}-\frac{1}{2}g\cdot \frac{n^2u^2}{g^2}$

$-H=\frac{nu^2}{g}-\frac{n^2u^2}{2g}$

Multiply by $g$ :

$-gH=nu^2-\frac{n^2u^2}{2}$

$gH=\frac{n^2u^2}{2}-nu^2$

$gH=u^2\left(\frac{n^2-2n}{2}\right)$

$gH=\frac{u^2n(n-2)}{2}$

Therefore,

$2gH=nu^2(n-2)$

This matches Option C.