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Q.

The coach throws a baseball to a player with an initial speed of 20ms-1  at an angle of 450 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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a

52 away from coach

b

52 towards the coach

c

25  towards the coach

d

25 away from the coach

answer is B.

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Detailed Solution

Rball =u2sin2θg = 202 sin 90010 = 40 m Time of flight  T =2usin θg=2×20×sin 45010  s = 22 s Therefore ×22 = 50 - 40  ⇒v = 52m/s
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