Question

A stone is dropped from rest into a well.It is observed to hit the water after 2s.Find the distance down to the water surface.How fast must the stone be thrown downwards in order to hit the surface after only 1s?what are the impact velocities in the two cases?

Answer 1

Question

From rest: $\nu_{0} = 0$ . Time falling: $t = 2$ s. The impact velocity: $\nu = \nu_{0} + g \cdot t = 0 + 9.8 \cdot 2 = 19.6 \, \mathrm{m} \cdot \mathrm{s}^{-1}$ .

The height of a well: $h = \nu_0 \cdot t + \frac{g \cdot t^2}{2} = 0 \cdot 2 + \frac{9.8 \cdot 4}{2} = 19.6 \, \mathrm{m}$ .

If we need time $t = 1$ s: $h = \nu_0 \cdot t + \frac{g \cdot t^2}{2} = 19.6 \, \mathrm{m} \Rightarrow \nu_0 = \frac{h - \frac{g \cdot t^2}{2}}{t} = \frac{19.6 - \frac{9.8 \cdot 1}{2}}{1} = 14.7 \, \mathrm{m} \cdot \mathrm{s}^{-1}$ .

The impact velocity: $\nu = \nu_{0} + g\cdot t = 14.7 + 9.8\cdot 1 = 24.5\mathrm{m}\cdot \mathrm{s}^{-1}$

Answer: the distance down to the water is 19.6 meters. The stone must be thrown downwards with the speed $14.7 \, \text{m} \cdot \text{s}^{-1}$ in order to hit the surface of the water after 1 second. The impact velocity in two cases are $19.6 \, \text{m} \cdot \text{s}^{-1}$ and $24.5 \, \text{m} \cdot \text{s}^{-1}$ .