Asked by anthony
a golf ball released from a height of 1.5 m above aa concreate floor bounces back to a height of 1.1m if the ball is in contact with the floor for .62 ms what is the average acceleartion of the ball while in contact with the floor?
Answers
Answered by
bobpursley
Vf^2=2ad
so in each of these, 1.5 meter, 1.1m find the floor velocity involved. Notice that they are opposite directions. a is g.
Now, having those two velocities, V1, V2
a=(V2-V1)/t but working in just speeds, and knowing they are in opposite directions
a=(V2+V1)/t
so in each of these, 1.5 meter, 1.1m find the floor velocity involved. Notice that they are opposite directions. a is g.
Now, having those two velocities, V1, V2
a=(V2-V1)/t but working in just speeds, and knowing they are in opposite directions
a=(V2+V1)/t
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