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Question: A body is thrown horizontally with a velocity \(\sqrt{2gh}\)from the top of a tower of height h. It ...

A body is thrown horizontally with a velocity 2gh\sqrt{2gh}from the top of a tower of height h. It strikes the level ground through the foot of the tower at a distance x from the tower. The value of x is –

A

h

B

h2\frac{h}{2}

C

2h

D

2h3\frac{2h}{3}

Answer

2h

Explanation

Solution

x = v2hgv\sqrt{\frac{2h}{g}}= 2gh\sqrt{2gh}× 2hg\sqrt{\frac{2h}{g}}= 2h.