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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.

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The ball continues to move horizontally with (Vx) = 10 m/sec. It begins to move up with `(V_y)_0` = 5 m/sec.. Therefore`theta_0`, is given as, `theta_0`. =` tan^-1(v_(y0))/(V_(x0))`
`implies theta = tan^-1` (5/10) = `tan^-1 (1/2)`.
Now the required radius of curvature is given as
`r=(^(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 `(/2)
we obtain r `cong` 14 m,
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