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(i) Let H be the height of the tower <br> The time of length, `T_(f) = sqrt((2H)/(g)) = 3s` <br> `rarr H = (g xx (3)^(2))/(2) = (9.8 xx 9)/(2) = 44.1 m` <br> (ii) Let the speed of projection be `v_(0)`. <br> Then for horizontal projection <br> `v_(x) = v_(0) rArr v_(y) =- g t` <br> At `t = T_(f) = 3s, v_(y) = - 9.8 xx 3 =- 29.4 ms^(-1)` <br> The angle which the final velocity makes with the horizontal `= theta = 45^(@)` (Given) <br> `rArr - tan 45^(@) = (-v_(y))/(v_(x)) rArr v_(y) = v_(x)` <br> So, `v_(x) = 29.4 ms^(-1)`