What Is the Radius of Curvature for a Horizontally Thrown Stone After 3 Seconds?

[lydiazmi]

Homework Statement

A stone was thrown horizontally with velocity 10 m/s. Find the radius of curvature of the stone trajectory 3 seconds after it was thrown.

Homework Equations

v²=v_x²+v_y²

a_n=(v²)/R

The Attempt at a Solution

v_x=10m/s
v_y0=0m/s
v_y=v_y0+gt=(9.8)(3)=29.4

v²=v_x²+v_y²=
=10²+29.4²
v=31.05m/s

now I've got the velocity, but according to the equ a_n=(v²)/R I hv to know the normal acceleration to find the radius. how??

(the answer is R=305m)

 

[gneill]

You have the components of the velocity vector at time t = 3 seconds, so you know the angle it makes with the horizontal. Pretend that defines a sloped surface with the stone on it. How would you decompose the gravitational acceleration vector into its surface normal and surface parallel components if this really were a block-on-a-slope problem?

 

[ashishsinghal]

For radius of curvature you need acceleration normal to the velocity. You know that only acceleration acting is g. Find the component of g normal to the velocity.

 

[lydiazmi]

great! thanks!