When kicking a football, the kicker rotates his leg about the hip joint. (a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity? (b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s? (c) Find the maximum range of the football, neglecting air resistance.

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aristocles aristocles · Answer 1 · 2019-10-29T05:13:24+01:00

Answer:

Part a)

[tex]\omega = 33.33 rad/s[/tex]

Part b)

[tex]F = 500 N[/tex]

Part c)

[tex]R_{max} = 40.8 m[/tex]

Explanation:

Part a)

As we know that the velocity of the tip of the kicker's shoe is given as

[tex]v = 35 m/s[/tex]

also the length of the tip of the shoe from his hip joint is given as

[tex]L = 1.05 m[/tex]

now the angular speed is given as

[tex]\omega = \frac{v}{L}[/tex]

[tex]\omega = \frac{35}{1.05}[/tex]

[tex]\omega = 33.33 rad/s[/tex]

Part b)

As we know that force on the ball is given as rate of change in momentum of the ball

so it is given as

[tex]F = \frac{\Delta P}{\Delta t}[/tex]

so we have

[tex]F = \frac{m(v_f - v_i)}{\Delta t}[/tex]

[tex]F = \frac{0.500(20 - 0)}{20 \times 10^{-3}}[/tex]

[tex]F = 500 N[/tex]

Part c)

As we know that the formula of range is given as

[tex]R = \frac{v^2 sin(2\theta)}{g}[/tex]

now for maximum range we know

[tex]\theta = 45 degree[/tex]

[tex]R_{max} = \frac{v^2}{g}[/tex]

[tex]R_{max} = \frac{20^2}{9.8}[/tex]

[tex]R_{max} = 40.8 m[/tex]