A shot putter launches a 7.120 kg shot by pushing it along a straight line of le

Question

A shot putter launches a 7.120 kg shot by pushing it along a straight line of length 1.650 m and at an angle of 34.80° 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 34.80°, and it lands at a horizontal distance of 13.50 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

 Mass of the shot putter is  m  = 7.120 kg  Horizontal distance  x = 13.5 m   Initial speed is v_i  = 2.50 m/s  Length  is   L  = 1.650m           t  = 13.5 / v cos 34.8      2.11  = x_0 sin 34.8 t - 1/2 gt^2      2.11 = 13.45 tan 34.8  - 1/2 (9.8)(13.5 /v cos 34.8)^2                v  = .........m/s          From Newton's law     F_net = F - mg sin 34.8 Now from work energy theorem       (F - mgsin34.5) *L = 1/2 m(vf^2 - vi^2)       F = mg sin 34.5  + [(1/2 ) m (v^2 - vi^2)] / L          solve for the above              

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