Jack and a Jill are standing on flat ground in front of an empty well that extends a depth s below the ground. Jill tosses a penny down the well with an initial velocity u0, while Jack decides to toss a penny up with an initial velocity v0.

How long does it take for Jack and Jill’s pennies to hit the bottom of the well? Express your answer in terms of know quantities such as v0, u0, s, and g, where g is the acceleration due to gravity.

To find the time it takes for Jack and Jill's pennies to hit the bottom of the well, we can use the equations of motion.

First, let's consider Jill's penny. We know that the well extends a depth s below the ground, so we can write the equation of motion for her penny as:

s = u0 * t + (1/2) * g * t^2

where t is the time it takes for the penny to hit the bottom of the well. Here, u0 is the initial velocity of Jill's penny and g is the acceleration due to gravity.

Next, let's consider Jack's penny. Since he tosses it up, the initial velocity v0 will be in the opposite direction to the gravity, so we can write the equation of motion for his penny as:

0 = v0 * t - (1/2) * g * t^2

where v0 is the initial velocity of Jack's penny.

Now, we can solve these two equations simultaneously to find the time when both pennies hit the bottom of the well.

Rearranging the equations, we have:

s - (1/2) * g * t^2 = u0 * t (Equation 1)
v0 * t - (1/2) * g * t^2 = 0 (Equation 2)

From Equation 2, we can solve for t in terms of v0 and g:

v0 * t = (1/2) * g * t^2
2 * v0 = g * t
t = (2 * v0) / g

Now, we substitute this value of t into Equation 1 to get:

s - (1/2) * g * ((2 * v0) / g)^2 = u0 * ((2 * v0) / g)

Simplifying further:

s - (1/2) * (4 * v0^2 / g) = (2 * u0 * v0) / g
s - (2 * v0^2 / g) = (2 * u0 * v0) / g
s = (2 * u0 * v0 + 2 * v0^2) / g

Therefore, the time it takes for both pennies to hit the bottom of the well is given by:

t = (2 * v0) / g

And the expression for s in terms of u0, v0, and g is:

s = (2 * u0 * v0 + 2 * v0^2) / g