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A shot putter launches a 7.220 kg shot by pushing it along a straight line of le

ID: 2011910 • Letter: A

Question

A shot putter launches a 7.220 kg shot by pushing it along a straight line of length 1.650 m and at an angle of 34.60° from the horizontal, accelerating the shot to the launch speed from its initial speed of 2.500 m/s (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of 2.110 m and at an angle of 34.60°, and it lands at a horizontal distance of 16.10 m. What is the magnitude of the athlete's average force on the shot during the acceleration phase? (Hint: Treat the motion during the acceleration phase as though it were along a ramp at the given angle.)
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Explanation / Answer

 Mass of the shot putter is  m  = 7.220 kg  Horizontal distance  x = 16.1 m   Initial speed is v_i  = 2.50 m/s  Length  is   L  = 1.650m           t  = 16.1 / v cos 34.6      2.11  = x_0 sin 34.6 t - 1/2 gt^2      2.11 = 16.1 tan 34.6  - 1/2 (9.8)(16.1 /v0 cos 34.6)^2                v  = ............m/s      From Newton's law     F_net = F - mg sin 34.6 Now from work energy theorem       (F - mg sin34.6) *L = 1/2 m(vf^2 - v0^2)       F = mg sin 34.6  + [(1/2 ) m (v^2 - v0^2)] / L     solve  to get solution     
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