Motion in a Plane

Concept:The horizontal component of a projectile's velocity remains constant throughout its motion because there is no horizontal acceleration.Explanation:Let the initial speed be $u$ at an angle $60^\circ$ with the horizontal.The horizontal component is $u \cos 60^\circ$.At the later instant, speed is $20\ \mathrm{m/s}$ at an angle $45^\circ$, so horizontal component is $20 \cos 45^\circ$.Since horizontal velocity is unchanged:$u \cos 60^\circ = 20 \cos 45^\circ$$u \times \frac{1}{2} = 20 \times \frac{1}{\sqrt{2}}$$u = 20 \times \frac{2}{\sqrt{2}} = 20 \times \sqrt{2}$Thus, $u = 20\sqrt{2}\ \mathrm{m/s}$.Answer:$20\sqrt{2}\ \mathrm{m/s}$ — Option (B).