Question

The coach throws a baseball to a player with an initial speed of 20 m/s at an angle of ${45}^{\circ }$ with the horizontal. At the moment the ball is thrown, the player is 50 m from the coach. At what speed and in what direction must the player run to catch the ball at the same height at which it was released? $(g=10m/s^2)$

• A. $\frac{2}{\sqrt{3}}m{s}^{-1}$
• B. $\frac{2}{\sqrt{5}}m{s}^{-1}$
• C. $\frac{5}{\sqrt{2}}m{s}^{-1}$
• D. $\frac{5}{\sqrt{3}}m{s}^{-1}$

Answer 1

The correct option is C $\frac{5}{\sqrt{2}}m{s}^{-1}$

$R=\frac{u^{2}sin2\theta }{g}=\frac{{20}^{2}sin2\times {45}^0}{10}=40mT=\frac{2usin\theta }{g}=\frac{2\times 20\times sin{45}^0}{10}=2\sqrt{2}m/sSotheplayerhastorunwithaspeeedofv=\frac{50-40}{T}=\frac{10}{2\sqrt{2}}=\frac{5}{\sqrt{2}}m/s$