Let the angle of projection be θ and initialvelocity be u \(\displaystyle{m}{s}^{{-{1}}}\).

Initial vertical component of velocity \(\displaystyle={u}{\sin{{0}}}={24}{\sin{{0}}}\)

Horizontaal component of velocity \(\displaystyle={u}{\cos{{0}}}={24}{\cos{{0}}}\)

Eqn. for the vertical component: \(v_{v}^{2} = [(24\sin θ)^{2} +- (2) \times (- 9.8 m s^{-1}) \times (-20 m)]\)

\(=576 \sin^{2}\theta + 392\)

The final horizontal component: \(v_{h}^{2} = 576\cos^{2}\theta\)

K.E. when the 2 kg mass touches the ground =

\((\frac{1}{2}) m[v_{v}^{2} +v_{h}^{2}] = 0.5 \times (2) \times (576\sin^{2}\theta + 392 + 576\cos^{2}\theta)\)

= 968 J