A shot putter launches a 7.49 kg shot by pushing it along a straight line of len

Question

A shot putter launches a 7.49 kg shot by pushing it along a straight line of length 1.69 m and at an angle of 34.5° from the horizontal, accelerating the shot to the launch speed from its initial speed of 2.4 m/s (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of 2.27 m and at an angle of 34.5°, and it lands at a horizontal distance of 16.1 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 = 7.49 kg, , = 34.5°, h = 2.27 m, D = 16.1 m. What is the magnitude F of the athlete's average force on the shotduring the acceleration phase After the shot leaves the hand, initial velocity = u (unknown),horizontal displacement = D, vertical displacement = -h D = ucos*t, so t = D/(ucos) -h = usin*t - gt2/2 = Dtan -gD2/(2u2cos2) gD2/(2u2cos2) = Dtan +h u = (D/cos)[g/(2Dtan + 2h)] = 11.84m/s During the acceleration phase, displacement d = 1.69 m, initialvelocity v = 2.4 m/s, final velosity = u = 11.84 m/s acceleration = a = (u2 - v2)/(2d) = 39.77m/s2 net force (along the ramp) on the shot = F - mgsin = ma F = m(a + gsin) = 339 N

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