From the top of a building- a particle is projected up at an angle - - with horizontal and reaches the ground in time - t-1- A second particle projected down at an angle - - with horizontal and with same speed reaches the ground in time - t-2- If third particle projected horizontally with same speed reaches the ground in time - t-3- then- t - 3 - t - 1 t - 2 - t - 3 - t - 1 - t - 2 2- 2 - t - 3 - 1 t - 1 - 1 - t - 2- t - 3 - t - 1 - 2 - t - 2 - 2 -

The correct option is A t _ 3 =√( t _ 1 t _ 2 ) h=usinθ #160; t _ 1 1 / 2 g t _ 1 ^ 2 .......(1) h=usinθ #160; t _ 2 1 / 2 g t _ 2 ^ ...

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Question

From the top of a building, a particle is projected up at an angle $θ$ with horizontal and reaches the ground in time $t_{1}$ . A second particle projected down at an angle $θ$ with horizontal and with same speed reaches the ground in time $t_{2}$ . If third particle projected horizontally with same speed reaches the ground in time $t_{3}$ , then

t_{3} = \sqrt{t_{1} t_{2}}

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t_{3} = \frac{t_{1} + t_{2}}{2}

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\frac{2}{t_{3}} = \frac{1}{t_{1}} + \frac{1}{t_{2}}

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t_{3} = \sqrt{t_{1}^{2} + t_{2}^{2}}

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