Question

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.)

Answer 1

Someone please Help !!!!!!!!!!!

Answer 2

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

Answer 3

I don't think that's the right soluyion i have tried thia and i got it wrong on my assignment

Answer 4

Misprint Vy•t - a•t²/2 = 2.11.
The solution is correct. Check my calculations

Answer 5

You need to account for the force of gravity which also has to be overcome and which you ignored.