Hammer Throw Homework: Find Force Applied by Athlete

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The discussion focuses on calculating the force applied by an athlete during a hammer throw, given specific parameters such as mass, release height, and distance traveled. The athlete whirls a 4.00 kg hammer at a constant speed along a circular path before releasing it at a 40-degree angle. Initial calculations yield a velocity of 27.1 m/s and a net force of 1728 N, but these differ from textbook values of 26.9 m/s and 1702 N. Key points of confusion include the neglect of the release height affecting flight time and the need for accurate kinematic equations. The participant successfully resolves the issue after addressing these factors.
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Homework Statement



An athlete whirls a 4.00 kg hammer six or seven times around and then releases it. ALthough the purpose of whirling it around several times is to increase the hammer's speed, assume that just before the hammer is released, it moves at constant speed along a circular arc of radius 1.7 m. At the instant she releases the hammer, it is 1.0 m above the ground and its velocity is directed 40 degrees above the horizontal. The hammer lands a horizontal distance of 74.0 m away. What force does the athlete apply to the grip just before she releases it? ignore air resistance.

Homework Equations



Force = mv^2/r
Kinematics:
Vf - Vi=at
Change in horizontal displacement = horizontal velocity * t

The Attempt at a Solution


I tried finding the time when the hammer reaches it's highest point, so the Vf would be zero. This equation is for the vertical (y direction)
Vf - Vi =at
vsintheta/a =t

So the total time is the time it takes to get to its highest point * 2
2vsintheta/a=Total time

Then plugging into displacement equation:
change in X=(Vx)i * T so:
vcostheta * 2vsintheta/a
solving for v=27.1 m/s.

Netforce =mv^2/r= 4.00 * (27.1)^2/1.7 m= 1728 N

My book gets 26.9 m/s for the velocity and then 1702 N for the Force. What am I doing wrong?! :(
 
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What is the actual question?
 
Without seeing the actual question, this is going to be difficult, but a different value of g? Incorrect conversion of units?
 
sjb-2812 said:
Without seeing the actual question, this is going to be difficult, but a different value of g? Incorrect conversion of units?

Sorry! FOr some reason, the question didn't show up the first time.
 
Sorry! For some reason, the question didn't show up the first time.
 
For one, you seem to be neglecting the fact it's released 1 metre above the ground, so the time taken for flight is not twice the time it takes to reach a maximum height.
 
sjb-2812 said:
For one, you seem to be neglecting the fact it's released 1 metre above the ground, so the time taken for flight is not twice the time it takes to reach a maximum height.

Thank you for reminding me of that!

After quite a bit of algebra, I got it!
 

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