Normal and tangential components

[BertsGarage]

Homework Statement

an athlete releases a shot at an angle of 20 degrees and velocity of 16m/s.
What are the velocity and acceleration of the shot in terms of normal and tangential components when it is at the highest point of its trajectory?
What is the instantaneous radius of curvature of the shot's path when it is at the highest point of its trajectory?

Homework Equations

v = v*et = ds/dt*et
a = at*et + an*en
where at = dv/dt and an = v*dtheta/dt = v^2/rho
rho = (1 + (dy/dx)^2)^3/2 / |d^2y/dx^2|

The Attempt at a Solution

I'm having a hard time with understanding how to express the velocity and acceleration in terms of normal and tangential components...Is there even a normal component to velocity?This is what I'm thinking...When the shot reaches the highest point of its trajectory, the normal velocity would be 0m/s and its normal acceleration would be -9.8m/s^2. The tangential acceleration would be 0m/s^2 which would leave only the tangential velocity to be calculated... I calculated vt = vo*et = 16m/s*(cos20+sin20) = 20.5m/s. For the instantaneous rate of curvature rho = v^2/an = 20.5m/s / -9.8m/s^2 = -2.09m, which doesn't make any sense. Any help would be greatly appreciated

 

[kuruman]

This is what I'm thinking...When the shot reaches the highest point of its trajectory, the normal velocity would be 0m/s and its normal acceleration would be -9.8m/s^2. The tangential acceleration would be 0m/s^2 which would leave only the tangential velocity to be calculated...

All of the above is correct.

I calculated vt = vo*et = 16m/s*(cos20+sin20) = 20.5m/s.

Not correct. The tangential velocity at the top of the trajectory is parallel to the ground, i.e. it is the horizontal component of the velocity at the top of the trajectory. How does that compare with the initial horizontal component of the trajectory?

For the instantaneous rate of curvature rho = v^2/an = 20.5m/s / -9.8m/s^2 = -2.09m, which doesn't make any sense. Any help would be greatly appreciated

At the top of the trajectory the instantaneous velocity and acceleration are perpendicular. How are the speed and acceleration related to curvature when this happens?

 

[BertsGarage]

The difference between the initial horizontal velocity and the velocity at the top of the trajectory is the angle the shot is moving. Initially it is tossed at a 20 degree angle but at the top of the trajectory since it is parallel with the ground, the angle is 0? So would the velocity at the top of the trajectory be the same as the initial velocity?
Would the instantaneous rate of curvature be 0 then because the tangential velocity is parallel with the ground?

 

[aim1732]

The part about initial horizontal velocity being equal to velocity at top is correct.

But for curvature you would have to look at angle b/w velocity vector and acceleration vector. Why think it as being parallel to the ground?

 

[kuruman]

The difference between the initial horizontal velocity and the velocity at the top of the trajectory is the angle the shot is moving. Initially it is tossed at a 20 degree angle but at the top of the trajectory since it is parallel with the ground, the angle is 0? So would the velocity at the top of the trajectory be the same as the initial velocity?

Not really. If the velocity at the top of the trajectory is the same as the initial velocity, then the velocity is not changing which means that the acceleration must be zero which you know is not the case. What is it stays the same at the top of the trajectory?

Would the instantaneous rate of curvature be 0 then because the tangential velocity is parallel with the ground?

No. The parabolic path still has curvature at the apex. If an object moves in a circle instantaneously and has speed v, how is the speed related to the acceleration that points toward the center of the circle? What is the name of the acceleration component that points toward the center of the circle?

 

[aim1732]

The part about HORIZONTAL velocity being unchanged is important.