A cosmic ray travels $60km$ through the earth’s atmosphere in $400\mu s$ as measured by experimenters on the ground. How long does the journey take according to the cosmic ray?

We have given that, A cosmic ray travels $60km$through the earth’s atmosphere in $400\mu s,$as measured by experimenters on the ground.

The cosmic ray enters the atmosphere and the cosmic ray hits the ground are two separate events.

In the cosmic ray's frame, frame$S^{\text{'}}$these can both be measured with a single clock. As a result, in $S^{\text{'}}$the proper time interval localid="1649596902268" $∆t^{\text{'}}=∆\tau$exists between them. The time interval measured in the Earth's frame $S,$is $∆t=400\mu s.$

The following is the outcome of time dilation:

$∆\tau =\sqrt{1-{\beta }^2}∆t$

The cosmic ray's speed in frame $S$is simply:

$$\begin{gathered} v=\frac{∆L}{∆t} \\ v=\frac{60000}{4.00\times {10}^{-6}} \\ v=1.5\times {10}^{8}m/s \end{gathered}$$

From previous, we have to find value for $\beta :$

$$\begin{gathered} \beta =\frac{v}{c} \\ \beta =\frac{1.5\times {10}^{8}m/s}{3\times {10}^{8}m/s} \\ \beta =0.5 \end{gathered}$$

Thus, the time interval measured by cosmic ray is:

$$\begin{gathered} ∆\tau =\sqrt{1-{(0.5)}^2}(400\mu s) \\ ∆\tau =346\mu s \end{gathered}$$