A stone is thrown horizontally with a velocity of `10m//sec`. Find the radius of curvature of it’s trajectory at the end of `3 s` after motion began. `(g=10m//s^(2))`

Options

A. `10 sqrt10` m
B. `100 sqrt10` m
C. `sqrt10` m
D. 100 m

Answer

B

Explanation

Radius of curvature , `r = (v_(t)^(2))/(" g sin" alpha)`
`v = sqrt(v_(x)^(2) + v_(y)^(2)) = 10 sqrt10`
tan `alpha = (v_(x))/(v_(y)) = (1)/(3) , "sin" alpha = (1)/(sqrt10)`