A shot putter launches a 6.910 kg shot by pushing it along a straight line of length 1.650 m and at an angle of 33.80° 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 33.80°, and it lands at a horizontal distance of 15.10 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.)
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Vx•t = 15.1
V•cos(33.8) •t = 15.1
t=15.1/(V•cos33.8) (1)
Vy•t+ a•t²/2 = 2.11
V•sin(33.8)•t - 0.5•(9.8)•t²= -2.11 (2)
Plugging (1) into(2)
V•sin(33.2)( 15.1/(V•cos33.8) - 4.9•{15.1/(V•cos33.8) }2 = -2.11,
tan(33.8)•(15.1)- 4.9•(15.1/(V•cos33.8))2 = -2.11.
Solve for final speed V
V=11.5 m/s
Now that you have the final velocity you can now solve for the acceleration.
V ²= V₀ ²+ 2•a•x
a= (V ²- V₀ ² )/2•x =(11.5²-2.5²)/2•1.65 =38.18 m/s²
F=m•a=6.91•38.18=263.8 N
V•cos(33.8) •t = 15.1
t=15.1/(V•cos33.8) (1)
Vy•t+ a•t²/2 = 2.11
V•sin(33.8)•t - 0.5•(9.8)•t²= -2.11 (2)
Plugging (1) into(2)
V•sin(33.2)( 15.1/(V•cos33.8) - 4.9•{15.1/(V•cos33.8) }2 = -2.11,
tan(33.8)•(15.1)- 4.9•(15.1/(V•cos33.8))2 = -2.11.
Solve for final speed V
V=11.5 m/s
Now that you have the final velocity you can now solve for the acceleration.
V ²= V₀ ²+ 2•a•x
a= (V ²- V₀ ² )/2•x =(11.5²-2.5²)/2•1.65 =38.18 m/s²
F=m•a=6.91•38.18=263.8 N
I don't think that's the right soluyion i have tried thia and i got it wrong on my assignment
Misprint Vy•t - a•t²/2 = 2.11.
The solution is correct. Check my calculations
The solution is correct. Check my calculations
You need to account for the force of gravity which also has to be overcome and which you ignored.
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