Question

A ball is dropped from a height of 12m. On every successive bounce, the ball bounces to a height that is 2/3 of the previous height. Find the total vertical distance that the ball has travelled when it hits the ground for the 8th time?

Answer 1

Hint: In above given problem, there is a ball dropped from a vertical distance of \[12m\] such that it bounces back every time it hits the ground and reaches the \[\dfrac{2}{3}\] rd of the previous height. Therefore, there are motions in two directions, one that is downward due to gravity and the other that occurs due to the bounce of the ball in upward direction. The total vertical distance is the sum of both distances covered in the respective directions. We can use the formula of sum of a G.P. to calculate the total distance up to 8th time without doing calculation of each step.

Complete step-by-step answer:
Given that the ball is dropped from a height of \[12m\] .
The ball bounces every time it hits the ground by a fraction of \[\dfrac{2}{3}\] of the previous height.
That means the ball falls from \[12m\] and then bounces back to a height of \[12 \times \dfrac{2}{3} = 8m\] and then again bounces to \[8 \times \dfrac{2}{3} = \dfrac{{16}}{3}m\] and so on.
Therefore the successive heights are \[12,12 \cdot \dfrac{2}{3},12 \cdot \dfrac{2}{3} \cdot \dfrac{2}{3},...\] and so on.
Clearly, it is a form of a G.P. where the first term is \[12\] and the common ration is \[\dfrac{2}{3}\] .
i.e.
\[ \Rightarrow a = 12\]
And
\[ \Rightarrow r = \dfrac{2}{3}\]
Now we can use the formula of sum of \[n\] terms of a G.P. which is given by,
\[ \Rightarrow {S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\]
The ball touches the ground \[8\] times, so \[n = 8\]
Therefore, total downward distance is given by
\[ \Rightarrow \dfrac{{12\left( {{{\left( {\dfrac{2}{3}} \right)}^8} - 1} \right)}}{{\dfrac{2}{3} - 1}}\]
On calculating ,that gives
\[ \Rightarrow 12 \cdot \left( {\dfrac{{256}}{{6561}} - 1} \right) \cdot \left( { - 3} \right)\]
Or,
\[ \Rightarrow 12 \cdot \left( {0.03901 - 1} \right) \cdot \left( { - 3} \right)\]
That gives,
\[ \Rightarrow 12 \cdot \left( { - 0.96099} \right) \cdot \left( { - 3} \right)\]
We can write it as
\[ \Rightarrow 12 \cdot \left( {2.88297} \right)\]
Therefore,
\[ \Rightarrow 34.595m\]
Now, for the upward distance,
Since the ball bounces back \[7\] times and the first distance is \[8\] , hence \[n = 7\] and \[a = 8\]
So the upward distance is given by,
\[ \Rightarrow \dfrac{{8\left( {{{\left( {\dfrac{2}{3}} \right)}^7} - 1} \right)}}{{\dfrac{2}{3} - 1}}\]
On calculating, that gives,
\[ \Rightarrow 8 \cdot \left( {\dfrac{{128}}{{2187}} - 1} \right) \cdot \left( { - 3} \right)\]
Or,
\[ \Rightarrow 8 \cdot \left( {0.05852 - 1} \right) \cdot \left( { - 3} \right)\]
That gives,
\[ \Rightarrow 8 \cdot \left( { - 0.9415} \right) \cdot \left( { - 3} \right)\]
We can write is as,
\[ \Rightarrow 8 \cdot \left( {2.8245} \right)\]
Therefore,
\[ \Rightarrow 22.595m\]
Hence, the total vertical distance is given by the sum of upward and downward distances,
Therefore, the total vertical distance is,
\[ \Rightarrow 34.595 + 22.595m\]
i.e,
\[ \Rightarrow 57.19m\]
Hence, the vertical distance covered by the ball is \[57.19m\] .

Note: The ball at the height of 12m falls downwards due to the attraction force of gravity by earth, and then bounces back in upward direction due to Newton's third law of motion that every action has an equal and opposite reaction. The bounced back distance and direction also depends on the material and shape of the ball and a few other physical factors.