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A 58.0 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it is descending vertically with(b) If the ball is in contact with the players head for 19 ms, what is the average acceleration of the ball? a speed of 26.0 m/s. If the player was moving upward with a speed of 4.20 m/s just before impact. (a) What will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic? (Note that the force of gravity may be ignored during the brief collision time.)

m₁=58 kg, v₁₀=4.2 m/s,
m₂= 0.45 kg, v₂₀= 26 m/s

v₂={ 2m₁v₁₀ - (m₂-m₁)v₂₀}/(m₁+m₂) =
={ 2•58•4.2 - (0.45-58) •26}/(58+0.45)=
=33.9 m/s.
F=m•Δv/t = m•(v₂-v₁)/t =58{33.9-(-26)}19•10⁻³=1.82•10⁵ N.

(b) If the ball is in contact with the players head for 19 ms, what is the average acceleration of the ball?

F=m2•a

I think the value I am getting is too large, what value are you getting?

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