Question

You drop a rock into a well and count the seconds before you hear a splash. If you counted 10 seconds before you heard the splash, how deep is the well?

Answer 1

$t=t_1+t_2$

$t_1=\sqrt{\frac{2h}{g}}$

$t_2=\frac{h}{v_{sound}}=\frac{h}{340}$

$\sqrt{\frac{2h}{g}}+\frac{h}{340}=10\to h\approx385.6(m)$. Answer