An athlete whirls a 7.2-kg hammer six or seven times around and then releases it

Question

An athlete whirls a 7.2-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 7.9 m. At the instant she releases the hammer, it is 3.8 m above the ground and its velocity is directed 8.2o 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

Lets assume that that force = F = Ma so a=F /m

which is applied to the hammer in horizontal direction.

As per the question there were no vertical velocity given so we assume here it as zero (Uy = 0)

So we apply the basic kinematic equations here-

S = Uy .t + 1/2 at2

3.8 = 1/2(10) t2

t= 0.82 sec

It is the time taken by the hammer to reach the ground so it is also same for horizontal motion.

Than,

S = Ux.t + 1/2 at2

74 = 8.20 (0.82) + 1/2 (F/7.2) (0.82)2

F = 1440.77 N

Thank you !!! Have fun

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