JEE Main 2014 · 6 Apr · Shift 1 · Physics Q2

← JEE Main 2014 · 6 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 2 of 90

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 :

2gH=n^2u^2

gH=(n-2)^2u^2

2gH=nu^2(n-2)

gH=(n-2)u^2

Answer and worked solution

Official answer: C

Time to the top is $\frac ug$ , so the total flight time is $t=\frac{nu}{g}$ . With up positive, the ground is at $-H$ :
$-H=ut-\tfrac12gt^2=\frac{nu^2}{g}-\frac{n^2u^2}{2g}\implies2gH=nu^2(n-2).$

Topics:Motion in a straight line

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JEE Main 2014 · 6 Apr · Shift 1, Physics question 2