Motion with constant acceleration

[SpaceW]

Homework Statement

A person is standing over a well and drops a rock. After 2.4 seconds he hears a rock splashing into the water. Take the speed of sound to be 300m/s and calculate the depth of the well.

Homework Equations

x=v/t for constant velocity
x=1/2at^2 for accelerated motion

The Attempt at a Solution

I tried this solving by taking t1=x/v x is depth of well and v speed of sound and also t2=(2x/g)^1/2
t1+t2=2.4 but i keep getting x to be 104.56 meters which can't be right although it seems all my steps are conclusive. Substituted t1 and t2 took p=x^1/2 and solved quadratic equation for p and god 10.4 m which goes to 104 m = x which is not right. What do you think is wrong?

 

[CWatters]

As a check work the numbers backwards..

On the way down..

d = ut+ 0.5gt²
u=0 so..
t = SQRT(2d/g) = SQRT(2*104/9.81) = 4.6 seconds

So you made a mistake somewhere.

Show your working?

 

[SpaceW]

T1+T2=2.4
X/v + ((2x)^1/2)/g^1:2 = 2.4
(g^1/2)x + v(2^1/2)x^1/2 = c
Call c = 2.4vg^1/2
take p=x^1/2
(g^1/2)p^2 + v(2^1/2)p - c = 0
I get p = 10.4 m

 

[PeroK]

T1+T2=2.4
X/v + ((2x)^1/2)/g^1:2 = 2.4
(g^1/2)x + v(2^1/2)x^1/2 = c
Call c = 2.4vg^1/2
take p=x^1/2
(g^1/2)p^2 + v(2^1/2)p - c = 0 [*]
I get p = 10.4 m

I think your equation [*] is correct, but you made a mistake solving it.

 

[SpaceW]

I solved it twice getting the same answer

 

[PeroK]

I solved it twice getting the same answer

Maybe you made the same mistake twice. You didn't forget something in the denominator of the formula for quadratic roots by any chance?

 

[SpaceW]

Oops 2a, not a...