First slide

Projection Under uniform Acceleration

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 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:

Moderate

A

v
B

2v
C

$\sqrt{2} v$
D

$\frac{v}{2}$

Solution

For the first body, h = $\frac{1}{2} {gt}^{2}$ ----------(i)
and x = vt -----------(ii)
From equation (i) and (ii)
h = $\frac{1}{2} g . (\frac{x^{2}}{v^{2}}) - - - - - - - (iii)$
$2 h = \frac{1}{2} g . (\frac{4 x^{2}}{{(v')}^{2}}) - - - - - - - (iv)$
Dividing equation (iii) by equation (iv) we get
$\frac{1}{2} = \frac{x^{2}}{v^{2}} \times \frac{{(v')}^{2}}{4 x^{2}} or v^{'} = \sqrt{2} v$

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