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A particle is thrown horizontally from the top of a tall tower with a speed of 10m/s. if radius of curvature of path followed is `4sqrt(2k` m at t=1sec, then find the value of k.

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Correct Answer - 5
image
At t=1sec
`u=10hati-10hatj`
`R=(v^(2))/(a_(N))=((10sqrt(2))^(2))/(g cos 45^(@))`

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