A shot putter launches a 7.110 kg shot by pushing it along a straight line of le

Question

A shot putter launches a 7.110 kg shot by pushing it along a straight line of length 1.650 m and at an angle of 34.00° 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 34.00°, and it lands at a horizontal distance of 14.30 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

Horizontal distance x = 14.30 m Initial speed of shot  u   = 2.50 m/s Length of straight line is   L = 1.650 m Angle of projection = 34o Horizontal distance                x =   v cos . t               t =   x / v cos                  =   14.30 m /  v cos 34     ...... (1) From kinematic equations                   2.11 = x sin 34.8 t - 1/2 gt2         2.11 = (13.45) tan 34  - (1 / 2 )( 9.8 m/s2 )( 13.5  / v cos 34 )2             v 2 =    186.626             v   =   13.66 m/s     Acceleration of the shot is              a = v 2 - u 2 / 2 L                  = ( 13.66 ) 2 - ( 2.50 m/s ) 2 / 2 ( 1.650 m )                  = 54.65 m/s 2 Average force on the shot        Fnet = F - mg sin 34.8                  = m ( a - g sin )                   = ( 7.11 kg ) ( 54.65 m/s2 - ( 9.8 m/s 2 ) sin 34 )                   = 349.665 N     

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