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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

9 years ago

Answers : (4)

vikas askiitian expert
509 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


9 years ago
Black Widow
36 Points

y=xtanα-g*x2/2u2cos2 α

hence on solving you will get a as answer

9 years ago
Samuel Garry
61 Points
It's also is 8.2 metres that is a option as it is the most closest 6.6 metres rest I don't know
It can directly come by equation of trajectory
2 years ago
Anuj Tapare
13 Points
Take g= 9.8 so we can get ans 8.2 we have to take 9.8 due this is a real type que , and we all know that we assume 10 instead of 9.8
one year ago
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