Question #19885
a ball is dropped on the floor from a height of 10m.It rebounds to a height of 2.5m. If the ball is in contact with the floor for 0.01sec, the average acceleration during contact is.
Expert's answer
H=10m
H=gt^2/2
V1=gt=Sqrt(2*gh)=Sqrt200
H1=2.5m
h1=gt1^2/2
V2=Sqrt(2*gh1)=Sqrt50
(V1+V2)/t=a
a=2.1 km/s^2
H=gt^2/2
V1=gt=Sqrt(2*gh)=Sqrt200
H1=2.5m
h1=gt1^2/2
V2=Sqrt(2*gh1)=Sqrt50
(V1+V2)/t=a
a=2.1 km/s^2
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