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Projectile from a Height

A body of mas...

Question

A body of mass $m$ is projected horizontally with a velocity $v$ from the top of a tower of height $h$ and it reaches the ground at a distance $x$ from the foot of the tower. If a second body of mass $2 m$ is projected horizontally from the top of a tower of height $2 h$ , it reaches the ground at a distance $2 x$ from the foot of the tower. The horizontal velocity of the second body is

v

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2 v

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\sqrt{2} v

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\frac{v}{2}

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Solution

The correct option is C $\sqrt{2} v$
The horizontal distance travelled by a body projected from height $h$ with horizontal velocity $u$ is,

$x = u \sqrt{\frac{2 h}{g}}$

which is independent of the mass.

So,

$\frac{u_{1}}{u_{2}} = \frac{x_{1}}{x_{2}} \times \sqrt{\frac{h_{2}}{h_{1}}} = \frac{x}{2 x} \times \sqrt{\frac{2 h}{h}} = \frac{1}{\sqrt{2}}$

$($ Given: $x_{1} = x; x_{2} = 2 x; h_{1} = h; h_{2} = 2 h; u_{1} = v)$

Therefore, velocity of second body,

$u_{2} = u_{1} \sqrt{2} = \sqrt{2} v$

Hence, option $(C)$ is the correct answer.

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