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Pooja singh Grade: 11
        

Qno.41 in the exercises  of hc verma(i tried a lot but couldn't get it, so i w'll be very grateful if the sol. is explained properly)

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

here is the solution


 


boy throws the ball with velocity 9.8m/s, accleration of car is 1m/s2


after the throw ,ball will move in verticle direction to a certain maximum height...


it will take t time to reach to maximum height..


  applying   v =u -gt


 at maximum height velocity will become 0...


  so u =gt     0r


  t=u/g =1sec        (u =9.8 given)


now time taken by ball to reach the maximum height is  t...


time of fall is also t....


 so total time = T=2t=2sec


after two second ball is reached to ground and car has covered some distance during this period...


 distance covered by car is S=ut+at2 /2     


                    S=aT2 /2               (u is 0 ,a =1 )


                   S=2m ans

6 years ago
NIKHIL GARG
38 Points
										

1396_21215_Untitled.jpg

6 years ago
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