A cricketer of height 2.5 m throws a ball at an angle 30º with the horizontal such that it is received by another cricketer of same height standing at a distance of 50 m from the first one.The maximum height attained by the ball from the ground is

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3.6 m

6.1 m

7.2 m

9.7 m

answer is D.

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Detailed Solution

R = 50 mHR=14tan⁡θH=50413=7.2mfrom the ground height is 7.2 + 2.5 = 9.7 m

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