Equal Displacements Example

A sequence of balls are successively dropped from the top of a tower. The time interval between the release of consecutive balls is equal. At the precise instant that the ninth ball (ball 9) is released, the first ball hits the ground. Which of the balls in the sequence is at \(\frac{3}{4}\) of the height, \(h\), of the tower when the first ball hits the ground?

Step 1: Understand what is happening

A ball is dropped from a tower. After a time, \(t_i\), a second ball is dropped. After another time interval, \(t_i\), a third ball is dropped. This means that when ball 3 is dropped the total time that has passed is \(2t_i\).

After a third time interval, \(t_i\), a fourth ball is dropped. This means that at the instant that ball 4 is dropped the total time that has passed is \(3t_i\). This process continues until the ninth ball is dropped. At the same instant that the ninth ball is dropped, the first ball that was dropped strikes the ground. The ninth ball would have been released after 8 time intervals, therefore it took the first ball a time interval of \(8t_i\) to hit the ground.

Step 2: Describe the events mathematically

As this is a vertical motion problem where everything is falling towards the ground, it is simplest to choose downwards as the positive direction.

The first ball hits the ground when the ninth ball is dropped. This means that the first ball has fallen for a total time of \(( \text{9}- \text{1})t_i = \text{8}t_i\). The displacement of the first ball is also described by the equation \(\Delta \vec{x}=\vec{v}_{i}t+\frac{1}{2}\vec{g}t^2\). The balls are dropped so their initial velocity is zero, \(\vec{v}_{i}=0\) and the we know the time for the first ball to fall the height of the tower therefore, in the case of the first ball, we have:

\begin{align*} \Delta \vec{x} &=\vec{v}_{i}t+\frac{1}{2}\vec{g}t^2 \\ h &=\frac{1}{2}(\text{9,8})(\text{8}t_i)^2 \end{align*}

Let the \(n\)th ball be the ball that is at \(\frac{3}{4}\) of the height, \(h\), of the tower when the first ball strikes the ground. The \(n\)th ball will have fall for a time of \((n-1)t_i\). If it is \(\frac{3}{4}\) of the way up the tower it will have fall a distance of \(\frac{1}{4}h\). This motion is also described by \(\Delta \vec{x}=\vec{v}_{i}t+\frac{1}{2}\vec{g}t^2\), and therefore we know, in the case of the \(n\)th ball, we have:

\begin{align*} \Delta \vec{x} &=\vec{v}_{i}t+\frac{1}{2}\vec{g}t^2 \\ \frac{1}{4}h &=\frac{1}{2}(\text{9,8})((n-\text{1})t_i)^2 \\ h &=2(\text{9,8})((n-\text{1})t_i)^2 \end{align*}

We can't solve for \(n\) from this final equation because it will contain \(h\) and \(t_i\). We can use the two equations we have to eliminate \(h\). Both equations are written in the form \(h=...\) therefore we can equate them eliminating \(h\):

\begin{align*} \frac{1}{2}(\text{9,8})(\text{8}t_i)^2 & = 2(\text{9,8})((n-\text{1})t_i)^2 \\ \frac{1}{2}(\text{9,8})\text{8}^2t_i^2 & = 2(\text{9,8})(n-\text{1})^2t_i^2 \\ \left(\frac{2}{(\text{9,8})t_i^2}\right)\times\frac{1}{2}(\text{9,8})\text{8}^2t_i^2 & = \left(\frac{2}{(\text{9,8})t_i^2}\right)\times 2(\text{9,8})(n-\text{1})^2t_i^2 \\ \text{8}^2 & = 4(n-\text{1})^2 \\ \text{8} & = 2(n-\text{1}) \\ \text{4} & = (n-\text{1}) \\ n & = 5 \end{align*}

Step 3: Write the final answer

The fifth ball is at \(\frac{3}{4}\) of the height of the tower when the first ball strikes the ground.