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`