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 `2 sec`.
`V_(y)=u_(y)+ g t =-30 m//s`
and `V_(x)=10m//s :. V^(2)=V_(x)^(2)+V_(y)^(2)`
`rArr V= 10 sqrt(10) m//s`
Now, `tan alpha = (V_(x))/(V_(y))=(1)/(3)`
`rArr sin alpha =(1)/(sqrt(10))`

Radius of curvaute `r=(V_(_|_)^(2))/(g sin alpha)`
`r=100sqrt(10)m`
Method `(II)`
Let horizontal and vertical position of 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)=(dx)/(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 x=Vt =30m`
and `V=10m//s`
`:. r=100 sqrt (10)m`.