Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 3 (Q.No. 38)
38.
A ball is dropped from a height of 2.25 m on a smooth floor and rises to a height of 1.00 m after the bounce. The coefficient of restitution between the ball and the floor is
Discussion:
2 comments Page 1 of 1.
Mahadev said:
8 years ago
Here, V(velocity after seperation)=eU(velocity before impact).
(2gh)^1/2=(2gH)^1/2 *coefficient of restitution.
c.o.r(e)=(2g*1/2g*2.25)^1/2.
c.o.r(e)=0.44.
h=velocity with which ball impinges on the floor.
H=velocity with which ball rebounds.
(2gh)^1/2=(2gH)^1/2 *coefficient of restitution.
c.o.r(e)=(2g*1/2g*2.25)^1/2.
c.o.r(e)=0.44.
h=velocity with which ball impinges on the floor.
H=velocity with which ball rebounds.
Rakesh Dabhi said:
1 decade ago
C = (h/H)^1/2 = 0.67
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