?
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.
Answer & Solution
Correct Answer:
Option
C
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:
$$\eqalign{
& \frac{a}{{ {1 - r} }} \cr
& {\text{Now}}, \cr
& { {\frac{{120}}{{ {\frac{1}{5}} }}} + {\frac{{96}}{{ {\frac{1}{5}} }}} } \cr
& = 1080\,m \cr} $$
Join the Discussion
Login to post a comment or share your explanation.
LoginSo , (no.of terms + total terms)×120
=(4+5)×120
=1080
For the bounces upward,
S(ups) = (4/5)(120) / (1 - 4/5) = 480 m
Total = 1080 m
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