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Shivam Bhagat Grade: 9
        

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

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

horizontal component of velocity = Vx = Ucos@


            Ux = 25cos60=25/2 m/s


            Uy = 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 = Uyt - 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


 

6 years ago
Black Widow
36 Points
										

y=xtanα-g*x2/2u2cos2 α


hence on solving you will get a as answer

6 years ago
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