InterviewSolution
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InterviewSolution
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A stone projected with a velocity u at an angle q with the horizontal reaches maximum heights `H_(1)`. When it is projected with velocity u at an angle `(pi/2-theta)` with the horizontal, it reaches maximum height `H_(2)`. The relations between the horizontal range R of the projectile, `H_(1) and H_(2)`, isA. `R=4sqrt(H_(1)H_(2))`B. `R=4(H_(1)-H_(2))`C. `R=4(H_(1)+H_(2))`D. `R=(H_(1)^(2))/(H_(2)^(2))` |
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Answer» Correct Answer - a `H_(1)=(u^(2)sin^(2) theta)/(2g)` `H_(2)=(u^(2)sin^(2)(90^(@)- theta))/(2g)=(u^(2)cos^(2) theta)/(2g)` `H_(1)H_(2)=(u^(2)sin^(2) theta)/(2g)/(u^(2)cos^(2) theta)/(2g)` `=((u^(2)sin 2 theta)^(2))/(16g^(2))=(R^(2))/16` `:. R=4sqrt(H_(1)H_(2))` |
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