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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1 : 2

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1 : 1

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Solution

The correct option is C 1 : 1

For first particle: $v^{2} = u^{2} + 2 g h \Rightarrow v = \sqrt{u^{2} + 2 g h}$
For second particle: $v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{u^{2} + (\sqrt{2 g h})^{2}} = \sqrt{u^{2} + 2 g h}$
So the ratio of velocities will be 1 : 1.

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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

The correct option is C 1 : 1 For first particle: v2=u2+2gh⇒v=√u2+2gh For second particle: v=√v2x+v2y=√u2+(√2gh)2=√u2+2gh So the ratio of velocities will be 1 : 1.

The correct option is C 1 : 1

For first particle: $v^{2} = u^{2} + 2 g h \Rightarrow v = \sqrt{u^{2} + 2 g h}$
For second particle: $v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{u^{2} + (\sqrt{2 g h})^{2}} = \sqrt{u^{2} + 2 g h}$
So the ratio of velocities will be 1 : 1.