physics question

Question

a small pebble is thrown into a water well with initial velocity vo= 1.5m/s. the sound of the water splash is heard 2.0 after the pebble is released. calculate the depth of the welll. use 340 m/s for the speed of sound.

Answer 1

2 seconds is the time for the pebble to go down + the time for the sound to go up. We can write this as

 

2 = t_p + t_s

 

The distance that they both travel are equal, distance sound = distance pebble

d_s = d_p

 

Since distance = rate * time, we can write d_s as 340t_s

 

For d_p we can use our equation in one dimension and write d_p = v₀t_p - ¹/₂ a_gt² , or d_p = -1.5t_p-4.9t_p²

 

So now our equation d_s = d_p can be written as 

 

340t_s = -1.5t_p - 4.9t_p²

 

From our first equation, 2 = t_p + t_s. We can substitute t_s = t_p - 2 into the above equation and we get

 

340(t_p-2) = -1.5t_p - 4.9t_p²

 

Now you can find out what t_p equals and then find the depth of the well from that.