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## Wednesday, 3 August 2011

### Physics GRE - #2

Consider that a coin is dropped into a wishing well. You want to determine the depth of the well from the $T$ between releasing the coin and hearing it hit the bottom. Suppose $T=2.059 s$. What is the depth $h$ of the well?

• 20.77 m
• 19.60 m
• 23, 564 m
• 18.43 m
• 39.20 m
To find out the time it takes for a coin to drop to the bottom of the well, we know from equations of kinematics that $h=\frac{1}{2}gt^2$, where $g\approx 9.80 m/s^2$. Thus the time for the coin to drop to the bottom is $t_1=\sqrt{\frac{2h}{g}}$. The time for the sound to travel back up the well is $t_2=\frac{h}{v_{sound}}$. We want to solve for $h$, and we know that $t_1+t_2=T$ so we have $\sqrt{\frac{2h}{g}}+\frac{h}{v_{sound}}=T.$ Using $T=2.059 s$, $g=9.8 m/s^2$, $v_{sound}=340 m/s$, we obtain $h=19.6 m$ and $h=25400 m$ upon solving the equation. Obviously $h=19.6 m$ is the correct solution while the other solution is an extraneous one that we got in the process of solving for $h$.

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