Kinematics/speed of sound question

[nothingsus]

Homework Statement

A stone is dropped in a well. The splash is heard 3s later. What is the depth of the well?

Homework Equations

constant acceleration equations
take speed of sound to be 343 m/s.

The Attempt at a Solution

For the stone:
u = 0
a = g
s = ut + 0.5at^2

subbing it all in you get t_stone = +sqrt(2s/g)

For the sound wave:
u = v = 343 m/s
v = s/t
subbing it all in you get
t_wave = s/343

(same distance traveled for stone and wave = s)

so you have t_stone + t_wave = 3

therefore
+sqrt(2s/g) + s/343 = 3
squaring both sides and moving over the constant
(s^2/343^2) + 2s/g - 9 = 0

solving this quadratic I get s = 44.02m or -24054m

The answer is 40.7m. Where did I go wrong?

 

[TSny]

Check the step where you squared both sides of the equation. Note (a + b)² ≠ a² + b² in general.

It might be good to isolate the square root on one side of the equation before squaring the equation.

 

[nothingsus]

Check the step where you squared both sides of the equation. Note (a + b)² ≠ a² + b² in general.

It might be good to isolate the square root on one side of the equation before squaring the equation.

Oops. Nice spot. Thanks for the help!

 

[TSny]

OK. Another approach would be to let $x = \sqrt{s}$ and write your equation sqrt(2s/g) + s/343 = 3 as a quadratic equation in terms of $x$. You will then not have to square the equation. Solve for $x$ and then find $s$.