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The velocity at the maximum height of a projectile is half of its velocity of projection `u`. Its range on the horizontal plane isA. `(2u^(2))/(3g)`B. `(sqrt(3)u^(2))/(2g)`C. `(u^(2))/(3g)`D. `(u^(2))/(2g)` |
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Answer» Correct Answer - B At maximum height `v=u cos theta` `(u)/(2)=vrArrcos theta=(1)/(2)rArrtheta=60^(@)` `R=(u^(2)sin2theta)/(g)=(u^(2)sin(120^(@)))/(g)` `=(u^(2)cos30^(@))/(g)=(sqrt(3)u^(2))/(2g)` |
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