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April 23, 2014

April 23, 2014

Posted by **Lindsay** on Thursday, October 4, 2007 at 6:47pm.

Quidditch told me that the first steps would be to find the velocity as the ball first makes contact with the floor, and the velocity as the ball leaves the floor. But I'm still unsure how to do this...

- Physics -
**Quidditch**, Thursday, October 4, 2007 at 8:24pmthe velocity when the ball first contacts the floor is:

V1 = sqrt(2*g*s)

V1= sqrt(2 * 9.8(m/s^s) * 1.8m)

The velocity when the ball just leaves (rebounds) works the same way except for the distance sign

V2 = sqrt(2 * 9.8(m/s^2) * (-1.06m))

V1 - V2= delta V (you will be subtracting a negative number remember for V2.

It is given that delta t is 4.62ms

acceleration = (delta v) / (delta t)

Don't forget to change the ms time to seconds.

- Physics -
**Lindsay**, Thursday, October 4, 2007 at 8:40pmTHANK YOU!

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