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2.) In the mens hammer throw field event, athletes compete to throw a hammer as far as possible. A Hammer consists of a ball of mass 7.257kg attached to a cable of length 1.215 meters. Atheletes typicall spin the hammer 4 times before releasing. The world record for a hammer throw is 86.74 meters.

a) Assuming that the hammer is thrown at an angle of 45 degrees to the horizontal and neglecting air resistance, calculate the s peed of the ball when released for the world record throw

My solution to part a is as follows:

Vi = Initial velocity

Vf = Final velocity

The ball is launched at an angle of 45 degrees, therefore the y-component of the initial velocity is given by Vi sin (45) = y-component of initial velocity.

Since the y-component has a downward acceleration of -9.8m and it travels 86.74 meters and the final velocity is 0 m/s

then using the equation Vf^2 = Vi^2 + 2ad we get

(0)^2 = (Vi sin 45)^2 + 2(-9.8)(86.74)

1700.104 = (Vi sin 45)^2

Vi sin 45 = 41.2

Vi = 58.31m/s

Is this correct? Thanks.