A ball initially at rest, is dropped from building of height 10 m to ground as shown in the figure below. If the coefficient of restitution is e=0.7 and after collision, the ball rebounds back up to height h, find the value of h. (Take g=9.8 m/s2)
A
10 m
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B
4.9 m
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C
9.8 m
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D
None of these
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Solution
The correct option is B4.9 m Taking downward direction as +ve Just before collision with ground Velocity of the ball (→u1)=√2gh=√2×9.8×10=14 m/s
After collision, Let velocity of the ball (→v1)=−v m/s As ground is stationary ∴Velocity of the ground before collision u2 and after collision v2 will be 0 m/s.
From, coefficient of restiution (e)=→v2−→v1→u1−→u2 0.7=0−(−v)14−0 v=14×0.7=9.8 m/s
After rebound, at maximum height h, speed of the ball will become zero.
∴ According to law of conservation of energy, 12mv2=mgh h=v22g=(9.8)22×9.8 =4.9 m