First slide

Projection Under uniform Acceleration

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 velocities when they reach the earth's surface will be

Moderate

A

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

1:2
C

1:1
D

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

Solution

$f o r p a r t i c l e p r o j e c t e d v e r t i c a l l y d o w n w a r d s v^{2} = u^{2} + 2 g h f o r t h e p a r t i c l e p r o j e c t e d h o r i z o n t a l l y h o r i z o n t a l v e l o c i t y v_{x} = u a n d v e r t i c a l v e l o c i t y v_{y} a f t e r a h e i g h t h i s \sqrt{2 g h} t o l a l v e l o c i t y v = \sqrt{{(v_{x})}^{2} + {(v_{y})}^{2}} v^{2} = u^{2} + 2 g h$

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