A ball is hit with initial speed \(u \mathrm {~ms} ^ { - 1 }\), at an angle \(\theta\) above the horizontal, from a point at a height of \(h \mathrm {~m}\) above horizontal ground. The ball, which is modelled as a particle moving freely under gravity, hits the ground at a horizontal distance \(d \mathrm {~m}\) from the point of projection.
Prove that \(\frac { g d ^ { 2 } } { 2 u ^ { 2 } } \sec ^ { 2 } \theta - d \tan \theta - h = 0\).
Given further that \(u = 14 , h = 7\) and \(d = 14\), and assuming the result \(\sec ^ { 2 } \theta = 1 + \tan ^ { 2 } \theta\),