Answer to Question #91639 in Mechanics | Relativity for Marisol Cuevas

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 downward in order to hit the surface after only 1s? What are the impact velocities in the two cases?

Answer 1

1. To find the distance down to the water surface h we use the equation

$$-h={{v}_{0}}t-\frac{g{{t}^{2}}}{2}\,\,\,\,(1)$$

where v₀ is the initial velocity, g=10m/s² is the acceleration of gravity. Distance and acceleration are down, therefore negative. Substitute the known values, v₀=0 m/s and t=2 s

$$-h=\frac{10\,m/{{s}^{2}}\cdot {{\left( 2\,s \right)}^{2}}}{2}=-20\,m$$

So h=20 m.

2. Find the initial velocity v₀ at which the stone should be thrown down to hit the surface in 1s. Use the equation (1)

$${{v}_{0}}t=\frac{g{{t}^{2}}}{2}-h$$

or

$${{v}_{0}}=\frac{gt}{2}-\frac{h}{t}$$

Substitute the known values, t=1 s and h=20 m

$${{v}_{0}}=\frac{10\,m/{{s}^{2}}\cdot 1\,s}{2}-\frac{20\,m}{1\,s}=-15m/s$$

Velocity is down, therefore negative.

3. To find the impact velocities we use the equation

$$v={{v}_{0}}-gt$$

In the first case, v₀=0 m/s and t=2 s, we get

$${{v}_{1}}=0-10\,m/{{s}^{2}}\cdot 2\,s=-20\,m/s$$

In the second case, v₀= ‒15 m/s and t=1 s, we get,

$${{v}_{2}}=-15-10\,m/{{s}^{2}}\cdot \,1\,s=-25\,m/s$$

Velocities are down, therefore negative.