Radius of curvature of projectile path

[Dusty912]

Homework Statement

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.

Homework Equations

v_x=sin(α)*V
a_n=v_x²/ρ

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

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 : v_x=sin(20)*16
v_x=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

 

[TSny]

v_x=sin(20)*16

Check this.

 

[Dusty912]

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

 

[TSny]

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

Looks good.

 

[Dusty912]

thank you