A cricket player is capable of throwing a ball at velocity v. Neglecting air resistance, what angle from the horizontal should they throw at to achieve maximum distance before contact with the ground? How far is that distance?

Assume angle of throw theta. Resolving vertical and horizontal components: Vvert=v.sin(theta) and Vhoriz=v.cos(theta).From 'suvat' equations, time to vertical stationary t=v/a. This must be doubled to find the total airborne time. Using the speed above and acceleration being only gravity, we have t=2.v.sin(theta)/g.Horizontal speed remains constant, so we can find total horizontal distance travelled with s=ut=v.cos(theta).2.v.sin(theta)/g.Using the trig equality sin(2.theta)=2.sin(theta).cos(theta) that can be derived from a data book equation, we can simplify to s=v^2.sin(2.theta)/g.We wish to maximise this distance, however v and g are constant. Therefore we need to maximise sin(2.theta). This can be done by differentiation or simply by noting that the maximum of a sin graph is at 90deg. Therefore 2.theta=90deg, and theta=45deg.Knowing this, and substituting, we find that distance s=v^2/g.