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Q.

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

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a

$ν$

b

$v / 2$

c

$\sqrt{2} ν$

d

$2 v$

answer is C.

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Detailed Solution

For the first body
$h = \frac{1}{2} {gt}^{2} . . . (i)$
and $x = νt$ .....(ii)
From eqn. (i) and (ii),
$h = \frac{1}{2} g \cdot (\frac{x^{2}}{v^{2}}) . . . (iii)$
For second body, let $ν$ ' be the velocity of projection, then
$2 h = \frac{1}{2} g \cdot [\frac{(2 x)^{2}}{v^{' 2}}] . . . . (iv)$
Dividing eqn. (iii) by eqn. (iv), we get
$\frac{1}{2} = \frac{x^{2}}{v^{2}} \frac{v^{' 2}}{4 x^{2}} or v^{'} = \sqrt{2} v$

alternatively $R = u \sqrt{\frac{2 h}{g}}$

$\frac{R_{1}}{R_{2}} = \frac{u_{1}}{u_{2}} \sqrt{\frac{h_{1}}{h_{2}}}$

$\frac{R}{2 R} = \frac{v}{u_{2}} \sqrt{\frac{h}{2 h}}$

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

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