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A particle projected horizontally with $20 m / s$ from a tower of height $45 m$ then find after
what time velocity vector makes $45^{0}$ with horizontal ? $(T a k e g = 10 m / s^{2})$

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answer is 2.

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Detailed Solution

$\tan θ = \frac{V_{y}}{V_{x}} = \frac{g t}{20} \tan 45 = \frac{g t}{20}$

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