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A boy is standing inside a train moving ...

 A boy is standing inside a train moving with a constant velocity of magnitude 10 m/s. He throws a ball vertically up with a speed 5 m/s relative to the train. Find the radius of curvature of the path of the L ball just at the time of projection.

Text Solution

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The ball continues to move horizontally with `(v_(x))_(0)=10`m/sec. It begins to move up with `(v_(y))=5`m/sec.
Therefore `theta_(0)` is given as `theta_(0)=tan^(-1)((v_(y^(@)))/(v_(x^(2))))` or
`theta_(0)=tan^(-1)(5//10)=tan^(-1)(1//2)`

Now the required radius of curvature is give as `r=(v^(2))/(g cos theta)`
Putting `v=v_(0)=sqrt((v_(x))_(0)^(2)+(v_(y))_(0)^(2)),sqrt(10^(2)+5^(2))=5sqrt(5)`m/s
`g=10m//s^(2)` and
`theta=tan^(-1)(1//2)`
we obtain `r~=14m`
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