The hammer throw is a track-and-field event in which a 7.3 kgball is whirled aro
ID: 1666076 • Letter: T
Question
The hammer throw is a track-and-field event in which a 7.3 kgball is whirled around in a circle several times and then released. It then moves upward on the familiar curving path ofprojectile motion and returns to earth some distance away. The world record is 86.75 m. Assume that the ball iswhirled on a circle with radius 1.8 m and that its velocity at theinstant of release is directed 41° above the horizontal. Find the magnitude of the centripetal force acting on theball at the moment just prior to the moment of release.I don't understand how you deal with the angle of release in thisproblem. The expert answer says that you divide by sin 82,but I don't know why. Also how would the picture look forthis? The hammer throw is a track-and-field event in which a 7.3 kgball is whirled around in a circle several times and then released. It then moves upward on the familiar curving path ofprojectile motion and returns to earth some distance away. The world record is 86.75 m. Assume that the ball iswhirled on a circle with radius 1.8 m and that its velocity at theinstant of release is directed 41° above the horizontal. Find the magnitude of the centripetal force acting on theball at the moment just prior to the moment of release.
Explanation / Answer
x = 86.75 m, = 41o find v vertical displacement = vsint - gt2/2 = 0 so t = 2vsin/g horizontal displacement = x = vcost = vcos* 2vsin/g = v2*(2sincos)/g =v2sin(2)/g v = [gx/sin(2)] = [9.8*86.75/sin(2*41)]= 29.3 m/s
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