# Radius of curvature of projectile path

1. Apr 11, 2017

### Dusty912

1. The problem statement, all variables and given/known data
The athelete releases the shot with velocity v = 16 m/s at 20° above the horizontal. What is the instantaneous radius of curvature of the shot’s path when it is at the highest point of its trajectory? Enter an answer in meters up to the first decimal place. Use g = 9.81 m/s2.

2. Relevant equations
vx=sin(α)*V
an=vx2

where an is the normal acceleration, vxthe velocity is the x component of velocity at the height of the path and ρ is the radius of the curve

3. The attempt at a solution
so I found the x component of a initial velocity which is the velocity , at the top of the path. using : vx=sin(20)*16
vx=5.472322293m/s

then I used 9.81 as the the acceleration for the normal and used the second equation stated above to solve for ρ
9.81=((5.47322293)2)/ρ
ρ=0.327580256meters

2. Apr 11, 2017

### TSny

Check this.

3. Apr 11, 2017

### Dusty912

oops cosine instead of sine. how's the rest?

4. Apr 11, 2017

### TSny

Looks good.

5. Apr 11, 2017

thank you