In shot putting, many athletes elect to launch the shot at an angle that is smal
ID: 1335333 • Letter: I
Question
In shot putting, many athletes elect to launch the shot at an angle that is smaller than the theoretical one (about 42 degree) 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 7.45 kg shot is accelerated along a straight path of length 1.49 m by a constant applied force of magnitude 390 N, starting with an initial speed of 2.5 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 32 degree and 42 degree? By what percent is the launch speed decreased if the athlete increases the angle from 32 degree to 42 degree?Explanation / Answer
a)
ramp angle = 32°
The force that resists the applied force is the weight component of the ball.
Weight of ball = 7.45(g) = 73.01 N
Component of weight acting down incline = 73.01 sin 32 = 38.7 N
Net applied force acting up incline = 390 - 38.7 = 351.3 N
Net acceleration up incline = 351.3/7.45 = 47.2 m/s²
Speed of ball at end acceleration = Vf = [(Vi)² + 2ad] {where Vi = 2.5, a = 47.2, d = 1.49}
Vf = [(2.5)² + (2)(47.2)(1.49)] = 6.25 + 140.656 = 146.91 = 12.12 m/s
b)
ramp angle = 42°
Weight of ball = 7.45(g) = 73.01 N
Component of weight acting down incline = 73.01 sin 42 = 48.85 N
Net applied force acting up incline = 390 - 48.85 = 341.15 N
Net acceleration up incline = 341.15/7.45 = 45.79 m/s²
Speed of ball at end acceleration = Vf = [(Vi)² + 2ad] {where Vi = 2.5, a = 45.79, d = 1.49}
Vf = [(2.5)² + (2)(45.79)(1.49)] = 6.25 + 136.454 = 142.7042 = 11.95 m/s
c)
Decrease in speed = 12.12 - 11.95 = 0.17 m/s
Percentage decrease = (0.17/12.12) x 100 = 1.4 %
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