In shot putting, many athletes elect to launch the shot at an angle that is smal

Question

In shot putting, many athletes elect to launch the shot at an angle that is smaller than the theoretical one (about 42 degrees) at which the distance of a projected ball at the same speed and height is greatest. One reason has to do with the speed the athlete can give the shot during the acceleration phase of the throw. Assume that a 7.260 kg shot is accelerated along a straight path of length 1.650 m by a constant applied force of magnitude 380.0 N, starting with an initial speed of 2.500 m/s (due to the athlete's preliminary motion). What is the shot's speed at the end of the acceleration phase if the angle between the path and the horizontal is (a) 30.00 degrees and (b) 42.00 degrees? (Hint: treat the motion as though it were along a ramp at the given angle) (c) by what percent is the launch speed decreased if the athlete increases the angle from 30.00 degrees to 42.00 degrees?

Explanation / Answer

Let a is the acceleration of the shot. am=F-mgsin so a=F/m-gsin a1=47,44(m/s^2) a2=45,78(m/s^2) 2al=v^2-v0^2 so v^2=2al+v0^2 so v=sqrt(2al+v0^2) v1=12,76(m/s) v2=12,54 c) the increase is (12,76-12,54)/12,54=1,75%

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