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`        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)`
7 years ago

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
```
7 years ago
NIKHIL GARG
38 Points
```
```
7 years ago
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