RoyalCat
- 670
- 2
Homework Statement
A rock is dropped down a deep pit. The sound of its impact with the floor is heard after T seconds.
The speed of sound is a given, V.
Prove that the depth of the pit is:
H=\frac{(\sqrt{V^{2}+2gVT}-V)^{2}}{2g}
Homework Equations
Kinematic equations.
The Attempt at a Solution
T is the time it takes for two motions to transpire. The first is the stone falling, t_1. The second, is the sound traveling back up, t_2.
T=t_1+t_2
For the falling rock:
y(t)=H-\frac{g}{2}t^{2}
y(t_1)=0
\frac{g}{2}t_1^{2}=H
t_1=\sqrt{\frac{2H}{g}}
For the wave of sound:
y(t)=Vt
y(t_2)=H
Vt_2=H
t_2=\frac{H}{V}
T=\sqrt{\frac{2H}{g}}+\frac{H}{V}
Now all that remains, so it would seem, is to extract H, since we have one equation with just one variable, since T,V and g are all given.
T-\frac{H}{V}=\sqrt{\frac{2H}{g}}
T^{2}-\frac{2TH}{V}+\frac{H^{2}}{V^{2}}=\frac{2H}{g}
\frac{T^{2}V^{2}g}{V^{2}g}-\frac{2TVgH}{V^{2}g}+\frac{gH^{2}}{gV^{2}}=\frac{2V^{2}H}{gV^{2}}
T^{2}V^{2}g-2TVgH+gH^{2}=2V^{2}H
gH^{2}-(2TVg+2V^{2})H+T^{2}V^{2}g=0
H=\frac{2TVg+2V^{2}+\sqrt{(2TVg+2V^{2})^{2}-4T^{2}V^{2}g^{2}}}{2g}
And from here, the algebraic manipulation eludes me. Have I made a mistake somewhere along the way? I was thinking about completing the square or some other nonsense, but the coefficient of 2 before V^{2} looks like it's going to be a problem.