A ball is thrown at speed u = 10.0 m/s at an angle of 30.0 degrees to the ground at height, s = 0. How far does the ball travel horizontally from its starting position? (Ignore air resistance and taking g = 9.81 m/s^2)

First, find the speed of the ball in the horizontal (x) and vertical (y) directions. u_x = ucos(30) = 8.66m/s and u_y = 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 = u_y, a = -9.81, t = ?. Substitute these values in and rearrange gives 0 = u_yt - 0.5gt² . Factorise out a t where the solution for this t would be t = 0 and so can be ignored give 0 = u_y - 0.5gt. This can then be rearranged to give t = 2u_y/g, with the values from the question this give a time in the air of t = 1.02s. Then substitute this value into s=u_xt for the horizontal equation give s = 8.83m.