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3 years ago

14

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?

1 answer:

Taya2010 [7]3 years ago

8 0

Answer:

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 = $\sqrt{\frac{gx}{sin2\theta}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s$

Now,

The angular velocity can be calculated as:

$v = \omega R$

Thus

$\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s$

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