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A man can throw a stone with initial speed of `10m//s`. Find the maximum horizontal distance to which he can throw the stone in a room of height h m for: (i) `h = 2m` and (ii) `h = 4m`. |
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Answer» Correct Answer - (i) `4sqrt(6)` (ii) `10m` for maximum horizontal rnage `theta = 45^(@)` for `theta = 45^(@)` max. height achieved `= (u^(2)sin^(2) theta)/(2g) = (10^(2)sin6(2)45^(@))/(2xx10) = 2.5m` (i) Since height of room `= 2m` which is less than height found above `h = (u^(2)sin^(2)theta)/(2g) rArr 2 = (10^(2)sin^(2) theta)/(2xx10)` `rArr sin theta = sqrt((2)/(3)) & cos theta = sqrt((3)/(5))` `R = (u^(2)sin 2 theta)/(g) rArr R = (u^(2)2sin theta cos theta)/(g) = 4sqrt(6)m` (ii) `theta = 45^(@)` so `R = (u^(2)sin 2 theta)/(g)` `rArr R = (10^(2) xx sin 90^(@))/(g) rArr R = 10m` |
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