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 Answer: Option CShow Answer
Solution(By Apex 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:
$\begin{aligned}&\frac{a}{1-r}\\
&\text{ Now, }\\
&\frac{120}{\frac{1}{5}}+\frac{96}{\frac{1}{5}}\\
&=1080m\end{aligned}$
Alternate Solution: 120m as it was thrown down. $\begin{aligned}120\times\frac{4}{5}\end{aligned}$ = 96m on rebounding. 96m again on going downwards. $\begin{aligned}96\times\frac{4}{5}\end{aligned}$ = 76.8m on rebounding. 76.8m again on going downwards. $\begin{aligned}76.8\times\frac{4}{5}\end{aligned}$ = 61.4m in rebounding. 61.4m again on travelling downwards. And so on… We get a series from these observations Total distance travelled; $={120 + 96 + 96 + 76.8 + 76.8 + 61.4 + 61.4 + …}$ I can rewrite that as: $120 + 2\times(96 + 76.8 + 61.4 …)$ Or: $\begin{aligned}120 + 2\times(96 + 96\times\frac{4}{5} + 96\times\left(\frac{4}{5}\right)^2 + …)\end{aligned}$ Now, inside those brackets, we got ourselves a Geometric Progression or a GP with the first term a = 96 and the common ratio r = $\Large\frac{4}{5}$ = 0.8 As it is an infinite GP, the sum of all it’s terms is equal to $\Large\frac{a}{1-r}$ So, the sum of distances covered = $\begin{aligned}120 + 2\times\frac{96}{1–0.8} = {120} + {960}\end{aligned}$ = 1080m.
Alternate Solution: 120m as it was thrown down. $\begin{aligned}120\times\frac{4}{5}\end{aligned}$ = 96m on rebounding. 96m again on going downwards. $\begin{aligned}96\times\frac{4}{5}\end{aligned}$ = 76.8m on rebounding. 76.8m again on going downwards. $\begin{aligned}76.8\times\frac{4}{5}\end{aligned}$ = 61.4m in rebounding. 61.4m again on travelling downwards. And so on… We get a series from these observations Total distance travelled; $={120 + 96 + 96 + 76.8 + 76.8 + 61.4 + 61.4 + …}$ I can rewrite that as: $120 + 2\times(96 + 76.8 + 61.4 …)$ Or: $\begin{aligned}120 + 2\times(96 + 96\times\frac{4}{5} + 96\times\left(\frac{4}{5}\right)^2 + …)\end{aligned}$ Now, inside those brackets, we got ourselves a Geometric Progression or a GP with the first term a = 96 and the common ratio r = $\Large\frac{4}{5}$ = 0.8 As it is an infinite GP, the sum of all it’s terms is equal to $\Large\frac{a}{1-r}$ So, the sum of distances covered = $\begin{aligned}120 + 2\times\frac{96}{1–0.8} = {120} + {960}\end{aligned}$ = 1080m.
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