First slide

Projectile motion

Question

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 speeds when they reach the earth's surface will be

Moderate

A

$\sqrt{2 g h + u^{2}} : u$
B

$1 : 2$
C

$1 : 1$
D

$\sqrt{2 g h + u^{2}} : \sqrt{2 g h}$

Solution

When particle is thrown in vertical downward direction with velocity u, then the final velocity at the ground level is
$\begin{array}{l} v^{2} = u^{2} + 2 g h \\ ∴ v = \sqrt{u^{2} + 2 g h} \end{array}$
Another particle is thrown horizontally with same velocity then at the surface of earth.
Horizontal component of velocity $v_{x} = u$
Therefore, resultant velocity, $v = \sqrt{u^{2} + 2 g h}$
For both the particles, final speeds when they reach the earth's surface are equal.

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