A particle is projected on a horizontal ground and moves freely under gravity. The horizontal and vertical components of the initial velocity are 2u and u ms^1 respectively. The ball lands a distance 80m from the point of projection. Show that u=14 This is what I did: Taking motion in the horizontal plane------ 2u = 80/t Taking motion in the vertical plane----- displacement (s) =0 acceleration (a)= -9.8 initial velocity (u) = u time of motion (t) = t therefore using s=ut +(1/2)at^2 I get: 4.9t=u substituting this into the first equation should give me the correct value of u, but it doesnt. Can someone tell me where i went wrong?