An athlete whirls a 3.5-kg hammer six or seven times around and then releases it

Question

An athlete whirls a 3.5-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 6.5 m. At the instant she releases the hammer, it is 8.6 m above the ground and its velocity is directed 10.0o above the horizontal. The hammer lands a horizontal distance of 74 m away. What force does the athlete apply to the grip just before she releases it? Ignore air resistance.

Explanation / Answer

here

u = sqrt( x * g )

u = sqrt( 74 * 9.8 )

u = 26.9 m/s

then the net force is

F = m * v^2 / r

F = 3.5 * 26.9^2 / 6.5

F = 389.63 N

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