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  4. A bomb is fixed from a canon with a velocity of 1000 m/s making an angle of 30° with the horizontal (g=9.8 text m/s2) . Time taken by bomb to reach the highest point is:

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Q. A bomb is fixed from a canon with a velocity of 1000 m/s making an angle of $ 30{}^\circ $ with the horizontal $ (g=9.8\text{ }m/{{s}^{2}}) $ . Time taken by bomb to reach the highest point is:

KEAMKEAM 2000

Solution:

$ T=\frac{u\,\sin \theta }{g} $ Given, $ u=1000\text{ }m/s, $ $ \theta =30{}^\circ ,g=9.8m/{{s}^{2}} $ $ T=\frac{1000\times \sin 30{}^\circ }{9.8}=51\,s $