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An object is projected horizontally from a top of the tower of height h. The line joining the point of projection and point of striking on the ground makes an angle 45⁰ with ground,Then with what velocity the object strikes the ground

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$\sqrt{\frac{gh}{2}}$

$\sqrt{\frac{9 g h}{2}}$

$\sqrt{\frac{7 g h}{2}}$

$\sqrt{\frac{5 gh}{2}}$

answer is D.

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Detailed Solution

$Tan θ = \frac{h}{R}, R = u \sqrt{\frac{2 h}{g}} and θ = 45^{0} Therefore h = u \sqrt{\frac{2 h}{g}} \Rightarrow u = \sqrt{\frac{gh}{2}} horizontal velocity$

$Vertical velocity v_{y} = \sqrt{2 gh}$

$velocity while hits the ground = \sqrt{u^{2} + 2 gh}$

$= \sqrt{\frac{g h}{2} + 2 g h} = \sqrt{\frac{5 g h}{2}}$

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