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. `10sqrt(10)m`
B. `100sqrt(10)m`
C. `sqrt(10)m`
D. `100m`

Answer

B

Explanation

Method (I)
After 3 sec
`V_(y) = u_(y) + g t=-30 m//s`
And `V_(x) = 10 m//s`
`:. V_(2) = V_(x2) +V_(y2)`
`rArr V = 10 sqrt(10) m//s`
Now, `tan alpha =(V_(x))/(V_(y))=(1)/(3)`
`rArr sin alpha = (1)/(sqrt(10))`
Radius of curvature `r = (V_(bot)^(2))/(g sin alpha)`
`r=(V_(bot)^(2))/(g sin alpha) rArr r = 100 sqrt(10)m`
Method (II)
Let horizontal and vertical position at point P be x & y respectively.
`:. x = Vt`
And `y = (1)/(2) g t_(2)`
`:.` equation of trajectory `y = (gx^(2))/(2V^(2))`
`(dy)/(dx)=(gx)/(V^(2)) and (d^(2)y)/(dx^(2))=(g)/(V^(2))`
Radius of curvature
`r=({1+((dy)/(dx))^(2)}^(3//2))/((d^(2)y)/(dx^(2)))=((1+(g^(2)x^(2))/(V^(4)))^(3//2))/(g//v^(2))`
Now after 3s
`rArr x =Vt = 30 m`
and `V = 10 m//s`
`:. r = 100 sqrt(10) m`.