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 2 h, it reaches the ground at a distance 2 x from the foot of the tower. The horizontal velocity of second body is

Moderate

A

v
B

2 v
C

$\sqrt{2} v$
D

v/2

Solution

For first body, $h = \frac{1}{2} {gt}^{2}$ …(1)
and x = vt ….. (2)
From eqs. (1) and (2), $h = \frac{1}{2} g (x^{2} / v^{2})$ ….(3)
For second body, let v' be the velocity of projection, then
$2 h = \frac{1}{2} g \{\frac{(2 x)^{2}}{v^{' 2}}\}$ …. (4)
Dividing eq. (3) by eq. (4), we get
$\frac{1}{2} = \frac{x^{2}}{v^{2}} \times \frac{v^{' 2}}{4 x^{2}} or v^{'} = \sqrt{2} \cdot v$

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