Question

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the ball at an angle of 54.6 degrees above horizontal, and the ball travels a total horizontal distance of 30.1 m. What angular velocity must she have achieved (in radians/s) at the moment of the throw, assuming the ball is 1.15 m from the axis of rotation during the spin?

Answer 1

The angular velocity is 15.37 rad/s

Solution:

As per the question:

`\theta = 54.6^(\circ)`

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v =
`\sqrt{(gx)/(sin2\theta)}`

v =
`\sqrt{(9.8* 30.1)/(sin2* 54.6)}`

v =
`\sqrt{(9.8* 30.1)/(sin2* 54.6)}`

v =
`\sqrt{(294.98)/(sin109.2^(\circ))} = 17.67\ m/s`

Now,

The angular velocity can be calculated as:

`v = \omega R`

Thus

`\omega = (v)/(R) = (17.67)/(1.15) = 15.37\ rad/s`