Solving Hammer Launch Problem: 32.8m

[danago]

An athlete starts to swing a hammer in a 2m radius circle which is tilted up on one side. Each complete 360 degree swing takes 0.71s. The angle of the plane of the circle to the horizontal is 45 degrees.

The hammer is released at a height of 1.5m from the ground, at an angle of 45 degrees to the horizontal, while traveling at 18m/s. How far away does it land, assuming g=10ms^-2

I started by coming up with parametric equations, which give the vertical (y) and horizontal (x) displacement as a function of time (t).

$$x=18t sin(45)$$
$$y=18t cos(45) - 5t^2$$

Since it was released from a heigh of 1.5m off the ground, the ground corrosponds to a displacement of -1.5m. So i set y=-1.5 and solved for t, giving t=2.658s. Then using this value in my 'x' equation, i get a horizontal displacement of 33.8m.

Now, i checked the answer, and apparently it is 32.8m. Have i done something wrong, or is the 2 in the answer a typo?

 

[denverdoc]

danago, if i compute the speed of the hammer from the radius and period info, I get an answer that's a bit different than 18m/s but which more closely corresponds to the book answer. BTW, I believe you have sin and cos flipped in your eqns but since they are the same at 45 degrees, made no difference to the calculation.

 

[danago]

Thanks for that I don't know why they would state a new velocity in the question though, since it contradicts the other given information.

 

[denverdoc]

Yea, its weird. IIRC it was like 17.7, maybe they rounded up, but then should have been consistent throughout