÷ÈÓ°Ö±²¥

We have given that, A cosmic ray travels $60km$through the earth’s atmosphere in $400\mathrm{μ}s,$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" $∆t^{\text{'}}=∆\mathrm{τ}$exists between them. The time interval measured in the Earth's frame $S,$is $∆t=400\mathrm{μ}s.$

The following is the outcome of time dilation:

$∆\mathrm{τ}=\sqrt{1-{\mathrm{β}}^2}∆t$

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

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

From previous, we have to find value for $\mathrm{β}:$

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

Thus, the time interval measured by cosmic ray is:

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