A ball is thrown from the top of a tower with an initial velocity of 10 ms-1 at an angle of 30o with the horizontal. If it hits the ground at a distance of 17.3 m from the back of the tower, the height of the tower is (Take g: 10 ms-2)

Here, θ= 30o, u = 10 ms-1,           R = 17.3 m, g = 10 ms-2For horizontal motion, R = u cos θtt = Ru cos θ = 17.310 cos 300 = 17.3 ×210×3 = 17.3×210×1.73 = 2 sFor vertical motion, h = u sin θt = 12gt2 = 10 sin 300×2-12×10×22= 10-20 = -10 m.Height of tower = 10 m.