Asked by Courtney
A 55-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed of 29 m/s.
If the ball is in contact with the player's head for 16 ms, what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)
If the ball is in contact with the player's head for 16 ms, what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)
Answers
Answered by
Damon
I will assume he just stops the descent of the ball
change of ball velocity = 29 m/s
a = change of velocity/time
= 29/(16*10^-3) = 1.81 * 10^3 m/s^2
change of ball velocity = 29 m/s
a = change of velocity/time
= 29/(16*10^-3) = 1.81 * 10^3 m/s^2
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