A 80 kg soccer player jumps vertically upwards and heads the .45kg ball as it is

Question

A 80 kg soccer player jumps vertically upwards and heads the .45kg ball as it is descending vertically with a speed of 30 m/s. if the player is moving upwards with a speed of 4.0 m/s just before impact. what will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic? if the ball is in contact with the players head for 19ms, what is the average acceleration of the ball? (force of gravity may be ignored during brief collision)

Explanation / Answer

here we need to use the conservation of linear momentum :
Initial linear momentum = 0.45*(-30 j ) + 80*(4 j )
Final linear momentum = 80*v + 0.45*u
v = speed of the ball after the collision
u = speed of the ball after the collision
-13.5 j + 320 j = 80v j + 0.45u j
306.5 = 80v + 0.45u j = unit vector, in y axiss direction
When you have a elastic collision, then the value of e = 1
1 = - ( u - v) / ( -30 - 4)
34 = v - u
and : 306.5 = 80v + 0.45u
v = 4 m/s
u = -30 m/s
So, the speed of the ball after the collision will be : 4 m/s upward
and the speed of the person after collision is 30 m/s downward.
Using linear impulse : dp = Impulse
and also : Impulse = Force*time
Force*19*10^-3 = 0.45*( 4 + 30)
Force = 805.3 N
Force = mass*acceleration
acceleration = 1790 m/s^2

Hope that helps

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