WAEC Physics 2010 Paper

Concept:For a horizontal projection, the time of flight depends only on the vertical height, while the horizontal distance depends on the speed of projection.Explanation:The vertical motion starts from rest, so $u_y = 0$.Using $s = u_y t + \frac{1}{2}gt^2$ with $s = 45\,m$ and $g = 10\,ms^{-2}$:$45 = 0 + \frac{1}{2}(10)t^2$$45 = 5t^2$$t^2 = 9$$t = 3\,s$During this time, the body travels horizontally $30\,m$ with constant horizontal speed $u$.Horizontal distance $= u \times t$$30 = u \times 3$$u = 10\,ms^{-1}$Answer:The speed of projection is $10\,ms^{-1}$, which is option A.