A cricketer hits a ball with a velocity 25 m/s at 60° above the horizontal. How far above the ground it passed over a fielder 50 m from the bat (assume the ball is struck very close to the ground)

(a) 8.2m

(b) 9.0m

(c) 11.6m

(d) 12.7m

horizontal component of velocity = V_x = Ucos@

            U_x = 25cos60=25/2 m/s

            U_y = 25sin60 = 25(3)^1/2 /2

 it covers a distance of 50 m after time t        

since accleration along horizontal direction is 0 so

             V = d/t        

        25/2 = 50/t

            t = 4sec

now after 4 sec let the height attained by particle is H then

 H = U_yt - gt²/2

   = 25(3)^1/2*4/2 - g*16/2

   = 50(3)^1/2 - 80

   =6.6 m

u can get this result directly by using equation of trajectory ,

Y(height) = Xtan@[ 1- X/R ]               

R is horizontal range , X is horizontal distance from point of projection & @ is angle of projection

y=xtanα-g*x²/2u²cos² α

hence on solving you will get a as answer