Moustapha Jones drops a penny down a well to wish for good grades on his upcoming physics test. He times the penny's drop and he hears the splash exactly 2.53s after the release. Of sound travels at 332m/s, how deep is the well?

Respuesta :

Answer:

29.43m

Explanation:

Let's s be the depth of the well and g = 9.8 m/s2. According to the following equation motion we can estimate the time t it takes to drop the coin:

[tex]s = gt^2/2[/tex]

[tex]t^2 = 2s/g[/tex]

[tex]t = \sqrt{2s/g}[/tex]

Also the time [tex]t_s[/tex] it would take for the drop sound to travel a distance of s is

[tex] s = v_st_s[/tex]

where[tex]v_s[/tex] = 332m/s is the speed of sound

[tex]t_s = s/v_s = s/332[/tex]

We also have the total time from release to the splash hearing:

[tex] t + t_s = 2.53[/tex]

[tex] \sqrt{2s/g} + s/332 = 2.53[/tex]

[tex] \sqrt{2s/9.8} + s/332 = 2.53[/tex]

[tex] 0.45\sqrt{s} + s/332 = 2.53[/tex]

We can numerically solve this so s = 29.43 m