Question

steven competes in the hammer throw event for his high schools track and field team. the hammer, shown below, consists of a handle at the end of a chain, with a heavy metal ball at the other end of the chain. to throw the hammer, steven holds the handle, extends his arm, and rotates his body four complete times in a circle before releasing the hammer. the combined length of stevens arm holding the hammer and the hammer itself, forming the radius of the circle, is 6 1/4 ft. for one throw, at the moment the hammer is released, how far has the metal ball traveled?

a. 625π/16 ft
b. 50π ft
c. 625π/4 ft

Explanation:

1. First, recall the formula for the circumference of a circle:

• The formula for the circumference of a circle is \(C = 2\pi r\), where \(r\) is the radius of the circle.
• Here, the radius \(r=6\frac{1}{4}=\frac{25}{4}\) ft.
• Steven rotates the hammer - body system four complete times before releasing the hammer.

1. Then, calculate the total distance \(d\) the metal - ball travels:

• The distance traveled in one full rotation is the circumference of the circle with radius \(r\). So, the distance traveled in four full rotations is \(d = 4\times C\).
• Substitute \(C = 2\pi r\) into the equation for \(d\): \(d = 4\times2\pi r\).
• Replace \(r=\frac{25}{4}\) into the equation: \(d = 4\times2\pi\times\frac{25}{4}\).
• Simplify the right - hand side of the equation:
• First, \(4\times2\pi\times\frac{25}{4}=2\pi\times25 = 50\pi\) ft.

Answer:

B. \(50\pi\) ft