Hammer Throw Homework: Find Force Applied by Athlete

[sona1177]

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?! :(

 

[fss]

What is the actual question?

 

[sjb-2812]

Without seeing the actual question, this is going to be difficult, but a different value of g? Incorrect conversion of units?

 

[sona1177]

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.

 

[sona1177]

Sorry! For some reason, the question didn't show up the first time.

 

[sjb-2812]

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.

 

[sona1177]

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!