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If for the same range, the two heights attined are 20 m and 80 m, then the range will beA. 20 mB. 40 mC. 120mD. 160 m |
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Answer» Correct Answer - D According to given condition, `(h_(1))/(h_(2)) = (sin^(2)theta)/(sin^(2)(90^(@)-theta)) = (20)/(80)` `rArr tan^(2)theta = (1)/(4) rArr tan theta = (1)/(2)` `sin theta = (1)/(sqrt(5))` and `cos theta = (2)/(sqrt(5))` So, `h = (u^(2)sin^(2)theta)/(2g)` `rArr 20 = (u^(2))/(10g)` or `(u^(2))/(g) = 200` `:.` The range, `R = (u^(2) xx 2 sin theta cos theta)/(g) = (200 xx 2 xx 2)/(sqrt(5) xx sqrt(5))` `= (200 xx 4)/(5) = 160 m` |
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