A child drops a rock into a vertical mineshaft that is preci

A child drops a rock into a vertical mineshaft that is precisely 445.00 m deep. The sound of the rock hitting the bottom of the shaft is heard 10.89 s after the child drops the rock. What is the temperature of the air in the shaft? Assume the temperature is the same throughout the mine.

Solution

Depth h = 445 m

Time taken to hear the sound T = 10.89 s

We know T = t + t \'

Where t = time taken to reach the rock to mine

            t \' = time taken to reach the sound from mine to child

Initial velocity of the rock u = 0

Accleration a = g = 9.8m/s ²

Time taken to reach the mine t = t

Distance S = h =445 m

From the relation S = ut +(1/2)gt ²

                            h = (1/2)(9.8) t ²

                               = 4.9 t ²

                           t = (h/4.9) ^1/2

                             = (445/4.9) ^1/2

                             = 9.529 s

Therefore t \' = T - t

                    = 10.89 s - 9.529 s

                    = 1.36 s

Velocity of the sound in air v = h / t \'

                                              = 445 m / 1.36 s

                                              = 327.14 m/s

Velocity of the sound at 0 ^o C = 331.3 m/s

From the relation v = (331.3+0.606 t )

                      327.14 = (331.3+0.606 t )

                   0.606 t = 327.14 -331.3

                               = -4.094

                            t = -6.755 ^o C