Question
A stone in thrown vertically down into a well, the splash is heard after a time $t_{1}$. If the stone is thrown vertically up with the same speed from the same point, the splash is heard after a time $t_{2}$. The common initial speed (u) of the stone in both the cases is given by (the time taken by the splash sound to reach the observer is ignored).
Options
Solution
Sol. Key concept: Vertical motion with opposite initial velocities.
Take downward as positive and let h be the depth of the well. Ignore the sound-travel time as instructed. For the downward and upward throws respectively:
$$h = ut_{1} + \frac{1}{2}g{t_{1}}^{2},\ \ \ \ \ h = - ut_{2} + \frac{1}{2}g{t_{2}}^{2}$$
Equate the two expressions for h and factor the difference of squares.
$$u(t_{1} + t_{2}) = \frac{g}{2}(t_{2} - t_{1})(t_{2} + t_{1})$$
$$u = \frac{g}{2}(t_{2} - t_{1})$$
The upward throw takes longer, so t₂ > t₁ and u is positive. Hence (d).
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