A shot putter launches a 7.45 kg shot by pushing it along a straight line of len

Question

A shot putter launches a 7.45 kg shot by pushing it along a straight line of length 1.73 m and at an angle of 32.8° from the horizontal, accelerating the shot to the launch speed from its initial speed of 2.1 m/s (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of 2.10 m and at an angle of 32.8°, and it lands at a horizontal distance of 15.9 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.)

Explanation / Answer

m = 7.45kg vi = 2.1m/s X = 15.9m 15.9 = v0 cos32.8 t          t = 15.9 /v0 cos32.8 and        2.1 = v0 sin32.8 t - 1/2 gt^2         2.13 = 15.9 tan32.8 - 0.5(9.8)(15.9 /v0 cos32.8)^2
             v0 = 14.69m/s             Fnet = F - mgsin 36       (F - mgsin32.8) *L = 1/2 m(vf^2 - vi^2)       F = mgsin32.8 + [(0.5)m(vf^2 - vi^2)]/L          = (7.45)(9.8)sin32.8 + (0.5)(7.45)[(14.69)^2 - (2.1)^2] /1.73         = 39.55 + 455.1 = 494.65 N    m = 7.45kg vi = 2.1m/s X = 15.9m 15.9 = v0 cos32.8 t          t = 15.9 /v0 cos32.8 and        2.1 = v0 sin32.8 t - 1/2 gt^2         2.13 = 15.9 tan32.8 - 0.5(9.8)(15.9 /v0 cos32.8)^2              v0 = 14.69m/s             Fnet = F - mgsin 36       (F - mgsin32.8) *L = 1/2 m(vf^2 - vi^2)       F = mgsin32.8 + [(0.5)m(vf^2 - vi^2)]/L          = (7.45)(9.8)sin32.8 + (0.5)(7.45)[(14.69)^2 - (2.1)^2] /1.73         = 39.55 + 455.1 = 494.65 N   

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