First, find the speed of the ball in the horizontal (x) and vertical (y) directions. ux = ucos(30) = 8.66m/s and uy = usin(30) = 5 m/s. Using an appropriate suvat equation find the time until the ball lands back on the ground: s = ut + 0.5at2, s = 0, u = uy, a = -9.81, t = ?. Substitute these values in and rearrange gives 0 = uyt - 0.5gt2 . Factorise out a t where the solution for this t would be t = 0 and so can be ignored give 0 = uy - 0.5gt. This can then be rearranged to give t = 2uy/g, with the values from the question this give a time in the air of t = 1.02s. Then substitute this value into s=uxt for the horizontal equation give s = 8.83m.