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  4. A stone is thrown at an angle θ to the horizontal reaches a maximum height h. The time of flight of the stone is:

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Q. A stone is thrown at an angle $ \theta $ to the horizontal reaches a maximum height h. The time of flight of the stone is:

EAMCETEAMCET 1998

Solution:

Maximum height attained by projectile is $ h=\frac{{{u}^{2}}\sin \theta }{2g} $ ?(i) $ \therefore $ $ u\sin \theta =\sqrt{2gh} $ ?(ii) Time of flight is given by $ T=\frac{2u\sin \theta }{g} $ ?(iii) From Eqs. (ii) and (iii), we get the time of flight is $ =\frac{2\sqrt{2gh}}{g}=2\sqrt{\frac{2gh}{{{g}^{2}}}}=2\sqrt{\frac{2h}{g}} $