A pole-vaulter holds a $16\text{ft}$ pole, A barn has doors at both ends, $10\text{ft}$apart. The pole-vaulter on the outside of the barn begins running toward one of the open barn doors, holding the pole level in the direction he's running. When passing through the barn, the pole fits (barely) entirely within the barn all at once. (a) How fast is the pole-vaulter running? (b) According to whom-the pole-vaulter or an observer stationary in the barn--does the pole fit in all at once? (c) According to the other person, which occurs first the front end of the pole leaving the bam or the back end entering, and (d) what is the time interval between these two events?

The given data can be written as:

The length of pole is $L_0=16\text{ft}$.

The distance between both doors of barn is$L=10\text{ft}$.

In this problem formula of length contraction, time dilation and Lorentz transformations are used to find the speed of pole vaulter and time of event.

The speed of the pole vaulter is given as:

$L=L_0\sqrt{1-\frac{v^2}{c^2}}$

Here, is the speed of the light and its value is $3\times {10}^8\text{m}/\text{s}$.

Substitute all the values in the above equation.

$\begin{array}{rcl}10\text{ft}& =& \left(16\text{ft}\right)\sqrt{1-\frac{v^2}{c^2}}\\ \sqrt{1-\frac{v^2}{c^2}}& =& 0.625\\ 1-\frac{v^2}{c^2}& =& 0.391\\ v& =& 0.78c\\ & & \end{array}$

$$\begin{gathered} v=0.78\left(3\times {10}^8\text{m}/\text{s}\right) \\ v=2.34\times {10}^8\text{m}/\text{s} \end{gathered}$$

Therefore, the speed of pole-vaulter is $2.34\times {10}^8\text{m}/\text{s}$.

As the pole-vaulter is moving so he feels length contraction in the pole so he will never agree that pole fits all at once in door. The stationary observer in the barn observes that pole will fit all at once in door because he does not observe any length contraction.

Therefore, the stationary observer in the barn observes that pole will fit all at once.

The difference in the time for entering of back end and leaving of front end in the barn is given as:

$t_2-t_1=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left(-\frac{v}{c^2}L\right)$

Substitute all the values in the above equation.

$$\begin{gathered} t_2-t_1=\frac{1}{\sqrt{1-\frac{{\left(0.78c\right)}^2}{c^2}}}\left(-\frac{0.78c}{c^2}\left(10\text{ft}\right)\left(\frac{0.3048\text{m}}{1\text{ft}}\right)\right) \\ {t}_2-t_1=1.6\left(-\frac{0.78}{c\left(\frac{3\times {10}^8\text{m}/\text{s}}{c}\right)}\left(10\text{ft}\right)\left(\frac{0.3048\text{m}}{1\text{ft}}\right)\right) \\ {t}_2-t_1=-1.26\times {10}^{-8}\text{s} \end{gathered}$$

As the difference in time for entering of back end or leaving of front end is negative so front end leaves before back end enters in the barn.

Therefore, the front end leaves before back end enters in the barn.

The difference in the time for entering of back end and leaving of front end in the barn is given as:

$t_2-t_1=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left(-\frac{v}{c^2}L\right)$

Substitute all the values in the above equation.

$\begin{array}{rcl}{t}_2-t_1& =& \frac{1}{\sqrt{1-\frac{{\left(0.78c\right)}^2}{c^2}}}\left(-\frac{0.78c}{c^2}\left(10\text{ft}\right)\left(\frac{0.3048\text{m}}{1\text{ft}}\right)\right)\\ t_2-t_1& =& 1.6\left(-\frac{0.78}{c\left(\frac{3\times {10}^8\text{m}/\text{s}}{c}\right)}\left(10\text{ft}\right)\left(\frac{0.3048\text{m}}{1\text{ft}}\right)\right)\\ t_2-t_1& =& -1.26\times {10}^{-8}\text{s}\\ & & \end{array}$

As the difference in time for entering of back end or leaving of front end is negative so front end leaves before back end enters in the barn.

Therefore, the time interval between two events is $-1.26\times {10}^{-8}\text{s}$.