What is the peak velocity of the shot putter's weighted ball?

[PVnRT81]

Homework Statement

A shot putter throws her shot. At the moment the 4.00 kg weight leaves her hand, it as a velocity of 14.0 m/s and is 2.00 m above the ground. The weight reaches a maximum height of 6.50 m above the ground before falling back down.

What is the magnitude of the weighted ball's velocity at the peak of its trajectory?

Homework Equations

E_k = (0.5)(m)(v²)
E_p = (m)(g)(h)

The Attempt at a Solution

(0.5)(m)(v²) _initial + (m)(g)(h) _initial = (0.5)(m)(v²) _final+ (m)(g)(h) _final

(0.5)( 14 m/s)² + (2 m)(9.8 m/s²) = (0.5)(v²) + (9.8 m/s²)(6.5 m)

V_peak = 5.19 m/s

I don't know if this is correct... isn't the velocity at the peak supposed to be zero as all the energy is potential.

 

[PhysicoRaj]

isn't the velocity at the peak supposed to be zero as all the energy is potential.

Yes, but when? What type of motion are you dealing with?

 

[PhysicoRaj]

Hint: you mentioned a shot put.

 

[Doc Al]

V_peak = 5.19 m/s

Your approach is correct, but I don't see how you got that answer. Resolve for the velocity.

I don't know if this is correct... isn't the velocity at the peak supposed to be zero as all the energy is potential.

Note that the weight is not thrown vertically.

 

[PVnRT81]

The peak velocity is 10.38 m/s.

 

[Doc Al]

The peak velocity is 10.38 m/s.

That's better.