The components of the velocity are \(\displaystyle{v}_{{{o}{y}}}={v}_{{o}}{\sin{{50}}}={15}{\sin{{50}}}={11.5}\ \frac{{m}}{{s}}\)

\(\displaystyle{v}_{\otimes}={v}_{{o}}{\cos{{50}}}={9.64}\ \frac{{m}}{{s}}\) The time taken by the ball to reach the point P ' is

\(\displaystyle{S}={v}_{{o}}{y}{t}-{\left({\frac{{{1}}}{{{2}}}}\right)}{>}^{{2}}\)

\(\displaystyle{0}{r}{2.10}={11.5}{t}-{\left({\frac{{{1}}}{{{2}}}}\right)}{\left({9.8}\right)}{t}^{{2}}\)

\(\displaystyle{4.9}{t}^{{2}}-{11.5}{t}+{2.1}={0}\)

This Quadractic equation in t, is solved for the values of t

\(\displaystyle{t}={2.15}\ {s}\) and 0.2 s

as the opponent begins to move only after 0.3s later, we taket = 2.15s

Therefore, he catches the ball at time t'=(2.15-0.3)s

The total horizontal distance

\(\displaystyle{x}={v}_{\otimes}\cdot{t}\)

\(\displaystyle{\left({10}+{x}\right)}={v}_{\otimes}\cdot{2.15}{s}\)

\(\displaystyle={9.64}\cdot{2.15}'\)

The value of ' x ' is calculated.

Thus the required speed of the opponent \(\displaystyle{v}={\frac{{{x}}}{{{t}'}}}\)