A cricketer of height 2.5 ~m throws a ball at an angle of 30^ 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 is ( 30^ =0.577 )

Step-by-step solution

The height of both cricketers is h = 2.5 m . Since the ball is thrown and received at the same height, the vertical displacement is zero. The horizontal range R is given by R = u^2 (2 ) g . Given R = 50 m and = 30^ , we have 50 = u^2 (60^ ) g = u^2 3 2g . Thus, u^2 g = 100 3 . The maximum height H attained by the projectile above the point of projection is given by H = u^2 ^2 2g . Substituting the value of u^2 g : H = 1 2 ( u^2 g ) ^2(30^ ) = 1 2 100 3 ( 1 2 )^2 = 100 8 3 = 12.5 1.732 7.216 m . The total maximum he