# Homework Help: Simple problem. Depth of well providing sound reaches you in 1.5 second.

1. Apr 8, 2004

### new324

I was wondering you someone could help me with an alternative way of solving this problem. Just to help my algebra a bit.
You drop a rock down a well. Find the depth of the well providing the sound of the splash reaches you in 1.5 seconds.

First off, we know speed of sound is 343 m/s.
Here are my equations:
t1+t2=1.5 s t1 is time for rock to hit bottom t2 is time for sound to reach your ears.
x=.5*a*t^2 -> x=4.9*t1^2
v=x/t -> x=343*t2

Now I sub the last 2 equations to get:
343*t2=4.9*t1^2
Now we sub. for t2:
343*(1.5-t1)=4.9*t1^2
From here we can reduce, and solve with the quadratic formula.

I was wondering how you would solve the equation, if you sub. for t1 instead of t2. This would make the equation:
343*t2=4.9*(t2-1.5)^2

2. Apr 8, 2004

### Staff: Mentor

It's just another quadratic equation. Multiply out the term on the right (do the square) and put it into standard form.