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If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal isA. `tan^(-1)((H)/(R ))`B. `tan^(-1)((2H)/(R ))`C. `tan^(-1) ((4H)/(R ))`D. `tan^(-1) ((4H)/(H))` |
Answer» Correct Answer - C Maximum height, `H = (u^(2) sin^(2) theta)/(2g) " "…(i)` Horizontal range, `R = (u^(2) sin 2theta)/(g) = (2u^(2)sin theta costheta)/(g) " "...(ii)` Divide (i) by (ii), we get `(H)/(R ) = (tan theta)/(4) or tan theta = (4H)/(R ) or theta = tan^(-1) ((4H)/(R ))` |
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