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  4. For a projectile the ratio of maximum height reached to the square of flight time is: (g=10ms-2)

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Q. For a projectile the ratio of maximum height reached to the square of flight time is: $ (g=10m{{s}^{-2}}) $

EAMCETEAMCET 2000

Solution:

Height of projectile $ {{H}_{\max }}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g} $ Time of flight $ T=\frac{2u\sin \theta }{g} $ $ \therefore $ $ \frac{{{H}_{\max }}}{{{T}^{2}}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta /2g}{{{(2u\sin \theta /g)}^{2}}} $ $ =\frac{g}{g}=\frac{10}{8} $ $ \therefore $ $ {{H}_{\max }}:{{T}^{2}}=5:4 $