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(A) When a packet is dropped from an aeroplane moving with horizontal velocity $u$,its initial horizontal velocity is $u$ and its initial vertical vel

Vedclass · Accepted Answer

$\sqrt{u^2 + 2gh}$

Vedclass · Answer

(A) When a packet is dropped from an aeroplane moving with horizontal velocity $u$,its initial horizontal velocity is $u$ and its initial vertical velocity is $0$.As the packet falls through a height $h$,its horizontal velocity remains constant at $u$ (ignoring air resistance).The vertical velocity $v_y$ acquired by the packet upon reaching the ground can be calculated using the kinematic equation $v_y^2 = u_y^2 + 2gh$,where $u_y = 0$.So,$v_y = \sqrt{2gh}$.The resultant velocity $v$ at the earth's surface is the vector sum of the horizontal and vertical components: $v = \sqrt{v_x^2 + v_y^2}$.Substituting the values,$v = \sqrt{u^2 + (\sqrt{2gh})^2} = \sqrt{u^2 + 2gh}$.

An aeroplane is moving with horizontal velocity $u$ at height $h$. The velocity of a packet dropped from it on the earth's surface will be ($g$ is acceleration due to gravity).

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