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

A boy is standing inside a train moving with constant velocity of magn

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