Question

Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return. (a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in b)0000 s. (b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 33b)00 m/s in this well.

Calculate the distance to the water neglecting the time for sound to travel up the well.
a) d_{{water}} = frac{1}{2}gt_{{sound}}^2
b) d_{{water}} = v_st_{{sound}}
c) d_{{water}} = v_st_{{sound}} + frac{1}{2}gt_{{sound}}^2
d) d_{{water}} = v_st_{{sound}} - frac{1}{2}gt_{{sound}}^2

Answer 1

The distance to the water neglecting the time for sound to travel up the well is 19.6 m, and taking into account the time for sound to travel up the well is 664.00 m.

Step-by-step explanation:

The answer to part (a) can be calculated using the formula dwater = 1/2gtsound^2, where g is the acceleration due to gravity. Neglecting the time required for sound to travel up the well, the distance to the water is:

dwater = (1/2)(9.8 m/s2)(2.0000 s)2 = 19.6 m

In part (b), taking into account the time for sound to travel up the well, the formula to use is dwater = vsoundtsound. The distance to the water is:

dwater = (332.00 m/s)(2.0000 s) = 664.00 m