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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