A ball is thrown from a position \(O\) on level ground with initial velocity (in \(\mathrm{m\,s^{-1}}\)) \(\mathbf{u}=5\mathbf{i}+6\mathbf{j}\) where \(\mathbf{i}\) is a horizontal direction and \(\mathbf{j}\) is vertically upwards.

Taking gravity to be \(\quantity{10}{m\,s^{-2}}\) and ignoring air resistance, find how far from \(O\) the ball will land and find an equation for the trajectory of the ball.

sketch of the trajectory of the ball

We consider the horizontal and vertical motions separately. Horizontally, the velocity is constant at \(\quantity{5}{m\,s^{-1}}\) so we can write \[x=5t\] where \(t\) is the elapsed time in seconds.

Vertically, we have constant acceleration due to gravity. \[y=6t-\frac{1}{2}10t^2\]

The ball hits the ground when \(y=0\), which means \[6t-5t^2 = 0\] or \[t(6-5t) = 0\] which means that \[t = 0 \text{ or } \frac{6}{5}.\]

Substituting, we find \[x=0 \text{ or } 6\] so the ball lands \(\quantity{6}{m}\) from \(O\).

Eliminating \(t\) from the horizontal and vertical equations, we get \[y=\frac{6}{5}x-\frac{x^2}{5}\] which is the equation of the trajectory. It describes a parabola with a maximum at \(\left(3,\frac{9}{5}\right)\).

sketch of the trajectory with maximum distances labelled