A ball falls from a height of 10 m on to a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is 1/√2, what is the total distance travelled by the ball before it ceases to rebond?
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Use formula, total distance travelled by the the ball before it ceases to rebound, S = h(1 + e²)/(1 - e²)
[ it is derived by using Newton's coefficient of restitution concepts , as you know coefficient of restitution is ratio of velocity of separation to velocity of approach. ]
Where , e is coefficient of restitution , h is Initial height of ball from horizontal floor.
given, e = 1/√2 and h = 10m
so, total distance travelled by the ball before it ceases to rebound , S = 10(1 + 1/2)(1 - 1/2)
= 10(3/2)/(1/2)
= 30 m
Hence, answer is 30m
[ it is derived by using Newton's coefficient of restitution concepts , as you know coefficient of restitution is ratio of velocity of separation to velocity of approach. ]
Where , e is coefficient of restitution , h is Initial height of ball from horizontal floor.
given, e = 1/√2 and h = 10m
so, total distance travelled by the ball before it ceases to rebound , S = 10(1 + 1/2)(1 - 1/2)
= 10(3/2)/(1/2)
= 30 m
Hence, answer is 30m
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