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°) 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 7.12 kg shot is accelerated along a straight path of length 1.71 m by a constant applied force of magnitude 410 N, starting with an initial speed of 2.2 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)28° and (b) 42°? (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 28° to 42°?

Explanation / Answer

mass of the shotput m=7.12 kg

force aplied F=410 N

initial speed v1=2.2 m/sec

path distance d=1.71 m

final speed is v2

a)

by using law of conservation of energy,

K1+F*d=K2+m*g*h

(1/2*m*v1^2)+(F*d)=(1/2*m*v2^2)+(m*g*d*sin(theta))

if theta=28 degrees

(1/2*m*v1^2)+(F*d)=(1/2*m*v2^2)+(m*g*d*sin(theta))

(1/2*7.12*2.2^2)+(410*1.71)=(1/2*7.12*v2^2)+(7.12*9.8*1.71*sin(28))

====> v2=13.64 m/sec

final speed v2=13.639 m/sec

b)

if theta=42 degrees

(1/2*m*v1^2)+(F*d)=(1/2*m*v2^2)+(m*g*d*sin(theta))

(1/2*7.12*2.2^2)+(410*1.71)=(1/2*7.12*v2^2)+(7.12*9.8*1.71*sin(42))

====> v2=13.64 m/sec

final speed v2=13.392 m/sec

c)

the percent of decreased speed is =(13.639-13.392)/(13.639)

=(0.01811)

=1.81%

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