After striking the floor, a rubber ball rebounds to 4/5th of the height from which it has fallen. Find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 metres.
A. 540 m
B. 960 m
C. 1080 m
D. 1020 m
E. 1120 m
Answer: Option C
Solution (By Examveda Team)
The first drop is 120 metres. After this the ball will rise by 96 metres and fall by 96 metres. This process will continue in the form of infinite GP with common ratio 0.8 and first term 96. The required answer is given by the formula: $$\eqalign{ & \frac{a}{{ {1 - r} }} \cr & {\text{Now}}, \cr & { {\frac{{120}}{{ {\frac{1}{5}} }}} + {\frac{{96}}{{ {\frac{1}{5}} }}} } \cr & = 1080\,m \cr} $$This Question Belongs to Arithmetic Ability >> Progressions
4/5, 3/5,2/5,1/5 then next is rest.
So , (no.of terms + total terms)×120
=(4+5)×120
=1080
S(downs) = 120 / (1 - 4/5) = 600 m
For the bounces upward,
S(ups) = (4/5)(120) / (1 - 4/5) = 480 m
Total = 1080 m
A rubber ball is dropped from a height of 25m which strikes the ground & rebounds everytimes to the half of the height from where it falls down what is the total distance traveled by the ball to come rest position?
It means the ball looses 1/5 of its height i.e 20% from 1 to 4/5 then from 4/5 to (4/5 - (4/5 * 1/5)) = 16/25
So it forms a G.P Series of..Multiplied by 2 because it goes up and comes down
(120 +2*(120* 4/5 + 120*16/25 +120* 64/125 + ...+ till Infinity )
=> 120 + 2*120 * (4/5 )/ (1 - 4/5)
=>120 + 960 =1080m
Total distance travelled will be 1080m
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