Asked by unowen
If a ball is thrown at the height of 10m then it bounces by 3\5 of the 10 meter and continues to resting condition, what is the total distance covered by the ball until rest?
Answers
Answered by
Reiny
Make a sketch to see that:
distance covered between bounce 1 and 2 = 2(10)(3/5) <--- once up, then down
distance covered between bounce 2 and 3 = 2(10)(3/5)^2
distance covered between bounce 3 and 4 = 2(10)(3/5)^2
....
total distance = 10 + [2(10)(3/5) + 2(10)(3/5)^2 + 2(10)(3/5)^3 + .... ]
the part in [ .. ] is an infinite series with a = 20(3/5) and r = 3/5
distance = 10 + a/(1-r)
leaving the actual arithmetic to you, let me know what you get
distance covered between bounce 1 and 2 = 2(10)(3/5) <--- once up, then down
distance covered between bounce 2 and 3 = 2(10)(3/5)^2
distance covered between bounce 3 and 4 = 2(10)(3/5)^2
....
total distance = 10 + [2(10)(3/5) + 2(10)(3/5)^2 + 2(10)(3/5)^3 + .... ]
the part in [ .. ] is an infinite series with a = 20(3/5) and r = 3/5
distance = 10 + a/(1-r)
leaving the actual arithmetic to you, let me know what you get
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