A ball was thrown at velocity of 10 m/s at `37^(@)` to the horizontal. What is the rate of change of its speed when its radius of curvature is 51.2 m

Options

A. `5 m//s^(2)`
B. `5sqrt(3) m//s^(2)`
C. `6 m//s^(2)`
D. `8 m//s^(2)`

Answer

D

Explanation

`a_(N)=g cos theta=(v^(2))/R`
`v cos theta=10 cos 37^(@)`
`v=8/(cos theta)`
`g cos^(3) theta=64/51.2`
`cos^(3) theta=64/512`
`cos theta=1/2`
`theta=60^(@)`
`a_(t)=(dv)/(dt)=g sin theta=5sqrt(3) m//s^(2)`