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Question

(c) When particle thrown in vertical downward direction with velocity u then final velocity at the ground level

${v^2} = {u^2} + 2gh$ $	herefore$

Vedclass · Accepted Answer

$1:1$

Vedclass · Answer

(c) When particle thrown in vertical downward direction with velocity u then final velocity at the ground level

${v^2} = {u^2} + 2gh$ $	herefore$ $v = \sqrt {{u^2} + 2gh} $

Another particle is thrown horizontally with same velocity then at the surface of earth.

Horizontal component of velocity ${v_x} = u$

$	herefore$ Resultant velocity, $v = \sqrt {{u^2} + 2gh} $

For both the particle final velocities when they reach the earth's surface are equal.

A man standing on the roof of a house of height $h$ throws one particle vertically downwards and another particle horizontally with the same velocity $u$. The ratio of their velocities when they reach the earth's surface will be

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