13. + 1 points SerEssen 16P060. A 69 kg soccer player jumps vertically upwards a

Question

13. + 1 points SerEssen 16P060. A 69 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it is descending vertically with a speed of 24 m/s. If the player was moving upward 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? My Notes Ask Your Teacher m/s If the ball is in contact with the player's head for 20 ms, what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.) m/s2

Explanation / Answer

after collision velocity of player is v1 and of ball is v2 upward.

then for elastic collision,

velocity of approach = velocity of separation

24+ 4 = v2 - v1

v1 = v2 - 28

momentum conservation,

69 x 4 - 0.45 x 24 = 60 v1 + 0.45 v2

265.2 = 60v2 - 1628 + 0.45 v2

v2 = 32.2 m/s ........Ans

v2 = v0 + a t

32.2 = - 24 + a (20 x 10^-3)

a = 2809 m/s^2 ......Ans

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