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The coach throws a base ball to a player with an initial speed of 20 ms^–1 at an angle of 45^o with the horizontal. At the moment the ball is thrown, the player is 50 m from coach. The speed and the direction that the player has to run to catch the ball at the same height at which it was released in ms^–1 is

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5/√2 m/s away from coach

5/√2 m/s towards the coach

√2/5 m/s towards the coach

√2 / 5 m/s away from the coach

answer is B.

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Detailed Solution

Range $= \frac{(20)^{2} (1)}{10} = 40 m$

Time= $= \frac{2 \times 20 \times \frac{1}{\sqrt{2}}}{10} = 2 \sqrt{2} \sec$

Distance = 10 m

Hence ν = $\frac{10}{2 \sqrt{2}} = \frac{5}{\sqrt{2}} m / s towards the coach$

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