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

• A. $\sqrt{2gh + u^{2}}:u$
• B. 1 : 2
• C. 1 : 1
• D. $\sqrt{2gh + u^{2}}:\sqrt{2gh}$

Answer 1

Answer: (C)

1 : 1

Answer 2

For first particle : $v^{2} = u^{2} + 2gh$ ⇒ $v = \sqrt{u^{2} + 2gh}$

For second particle :

$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{u^{2} + \left( \sqrt{2gh} \right)^{2}} = \sqrt{u^{2} + 2gh}$

So the ratio of velocities will be 1 : 1.