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