First slide

Projection Under uniform Acceleration

Question

A stone is thrown horizontally with velocity g ms^-1 from the top of a tower of height g metre. The velocity with which it hits the ground is (in ms^–1)

Moderate

A

g
B

2g
C

√3g
D

4g

Solution

Just before the stone hits the ground, its horizontal component of velocity V_x = u = 'g' and vertical component of velocity is given by
$V_y^2 = 0 + 2.g.h \Rightarrow {V_y} = \sqrt {2gh} = \sqrt {2g.g} = \sqrt 2 g$
$∴$ Speed with which the stone hits the ground is
$V = \sqrt {V_x^2 + V_y^2} = \sqrt {{g^2} + 2{g^2}} = \sqrt 3 g$

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