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!

 

[rwisz]

I would first clarify the context of the problem. Is the stone being thrown on Earth, or in a vacuum? Is there any air resistance? These factors can affect the trajectory of the stone and its radius of curvature.

Assuming the stone is being thrown on Earth with no air resistance, we can use the equations of motion to solve for the radius of curvature. The stone is moving in a circular path, so its acceleration is directed towards the center of the circle and is equal to the centripetal acceleration. We can use the formula an = v^2/R to calculate this acceleration.

In this case, the velocity (v) is equal to 31.05 m/s, as calculated in the attempt at a solution. We can also calculate the centripetal acceleration (an) by using the formula an = (vy^2)/R, since the stone is moving horizontally with no vertical component to its velocity.

Therefore, we can set these two equations equal to each other and solve for R:

an = (vy^2)/R
an = (v^2)/R
(vy^2)/R = (v^2)/R
vy^2 = v^2
(29.4)^2 = (31.05)^2
R = (31.05)^2/(29.4)^2
R = 305 m

The radius of curvature of the stone's trajectory is 305 meters. It is important to note that this calculation is based on ideal conditions and may differ in a real-world scenario. Further analysis and experimentation may be needed to accurately determine the radius of curvature in a real-life situation.