• # question_answer 4) A ball is projected with a velocity of $10\sqrt{2}\,m/s$at angle of $45{}^\circ$ with the horizontal from the top of a cliff 20 m high, with what velocity does the ball hit the ground? A) $5\sqrt{3}\,m/s$                    B) $10\sqrt{3}\,m/s$C) $10\sqrt{6}\,m/s$                  D)  $10\sqrt{2}\,m/s$

Let the speed of projection be $u=10\sqrt{2}\,m/s$and the speed of the ball just before hitting the ground be v. Applying law of conservation of mechanical energy, we get: $\frac{1}{2}m{{u}^{2}}+mgh=\frac{1}{2}m{{v}^{2}}$ $\therefore$    $v=\sqrt{{{u}^{2}}+2gh}$ $=\sqrt{200+2\times 10\times 20}$ $=\sqrt{600}=10\sqrt{6}\,m/s$ Hence, the correction option is [c].