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Abody of mass in thrown up vertically with velocity v_(1) reaches a maximum height h_(1)in t_(1) seconds. Another body of mass 2 m is projected with a velocity v_(2) at an angle theta. The second body reaches a maximum height h_(2) in time t_(2) seconds. If t_(1) = 2t_(2), then ratio ((h_(1))/(h_(2)) is |
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Answer» 0.043055555555556 `h_(1) = (v_(1)^(2))/(2g)` Another body of mass 2m is projected with a velocity `v_(2)` at an angle `theta`. `therefore "Height attained" (h_(2)) = (v_(2)^(2)sin^(2)theta)/(2g)` `therefore "" (h_(1))/(h_(2)) = (v_(1)^(2))sin^(2) theta` `"But" "" t_(1) - 2t_(2)` `therefore "" (v_(2))/(g) = 2 [(v_(2)sintheta)/(g)]` `rArr "" v_(1) = 2v_(2)sintheta` `therefore "" (h_(1))/(h_(2)) = ((2)/(1))^(2) = (4)/(1)` |
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