While conceptually simple, drawing on understanding of kinematic equations and vectors, these projectile problems require attention to the details of the problem such as accounting for the launch height and taking care with the direction of acelleration. The launch should be solved using a vector diagram to separate the horizontal and vertical components of velocity (14.09 & 5.13 m/s). The time at which the ball changes direction is t=v/g=0.52 s and the maximum height comes from s=ut+0.5gt^2 (gives 1.34m travelled vertically + 1m = 2.34m). The falling section of the calculation follows from this, using the same kinematic equation but this time u=0 and solving for t (0.69s).Having solved the vertical components of the problem the distance travelled can simply be obtained using s=ut, remembering to use the total time of rising and falling (giving 17.05m).