Sam and Terry find a 29.0 m deep wishing well and decide to make a wish. Terry throws a penny down the well at 9.00 m/s and Sam throws a penny up in the air at the same speed. If they threw them at the same time, how much time would there be between splashes?

Terry:

d = Vo*t + 0.5g*t^2 = 29.
9t + 4.9t^2 = 29
4.9t^2 + 9t -29 = 0.
Use Quadratic Formula:
t = -9 +- Sqrt(9^2+4*4.9*29)/9.8 = 1.68 s. = Time to reach bottom of well.

Sam:
V = Vo + g*Tr = 0.
9 - 9.8Tr = 0
Tr = 0.918s = Rise time.
Tf1 = Tr = 0.918s = Fall time to launching point.

d = Vo*Tf2 + 0.5g*TF2 = 29.
9Tf2 + 4.9(Tf2)^2-29 = 0
Tf2 = 1.68 s.
T = Tr+Tf1+Tf2 = 0.918 + 0.918 + 1.68 = 3.52 s.

T-t = 3.52 - 1.68 = 1.84 s. Between splashes.

Well, well, well, it seems like Sam and Terry are in quite the penny predicament! Let's crunch some numbers and see how long their wishes will take. Since both pennies are thrown with the same initial speed, we know their initial velocities are both 9.00 m/s.

The penny thrown down the well will experience free fall, so we can use the equation h = v₁t + 0.5gt², where h is the depth of the well (29.0 m), v₁ is the initial velocity (9.00 m/s), g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time it takes to reach the bottom.

Using this equation, we rearrange it to solve for t: t = √((h - 0.5gt²) / v₁).

Substituting the given values, we get t = √((29.0 - 0.5 * 9.8 * t²) / 9.00).

Is it just me, or is this equation getting a bit complicated? Let's simplify things and make an estimation. Assuming the time it takes for the penny to reach the bottom of the well is relatively short compared to the time it takes for the thrown penny to reach its peak height and fall back down (since the latter has a higher distance to cover), we can neglect the distance traveled by the penny going down.

Therefore, we can focus on the time it takes for Sam's penny to reach its peak height and fall back down. The time it takes for an object thrown vertically to reach its peak height and return to the same height is equal to twice the time it takes to reach the peak.

So, we can approximate the time between splashes to be roughly 2 times the time it takes for the penny to reach its peak height. We can use the same kinematic equation h = v₀t + 0.5gt², where h is the height (let's say 29.0 m), v₀ is the initial velocity (9.00 m/s), g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time to reach the peak.

Plugging in the values, we get 29.0 = 9.00t - 0.5 * 9.8 * t².

Solving this equation, we find two possible solutions: t = 0.72 s or t = 3.92 s.

Since we're considering the time it takes for the penny to reach its peak height and return, we choose the smaller value, t = 0.72 s.

Therefore, if Sam and Terry throw their pennies at the same time, there would be approximately 2 * 0.72 = 1.44 seconds between their splashes.

Remember, this calculation is just an estimation, so it might not be 100% accurate, but it should give you a general idea. Keep the laughter flowing, my friend!

To find out how much time there would be between the splashes, we need to determine the time it takes for each penny to reach the bottom of the well.

Let's consider Sam's penny first. The initial speed of the penny is 9.00 m/s, and we need to find the time it takes for the penny to reach the bottom of the well, which is 29.0 m deep.

Using the equation of motion:

v^2 = u^2 + 2as

where,
v = final velocity (we take it as 0 m/s, as the penny would come to rest at the bottom of the well)
u = initial velocity (9.00 m/s)
a = acceleration (we consider it as -9.8 m/s^2, as the penny is moving against the gravitational force)
s = distance (29.0 m)

Rearranging the equation to solve for time:

t = (v - u) / (-a)

Substituting the values into the equation:

t = (0 - 9.00) / (-9.8)

Calculating the time for Sam's penny:

t = 9.00 / 9.8

t ≈ 0.9184 s

Now, let's find the time it takes for Terry's penny to reach the bottom of the well. Since Terry is throwing the penny upwards at the same speed, the initial velocity of the penny is also 9.00 m/s. The only difference is that the penny is moving against gravity in this case.

Using the same equation of motion:

t = (v - u) / (-a)

Substituting the values:

t = (0 - 9.00) / (-9.8)

Calculating the time for Terry's penny:

t = 9.00 / 9.8

t ≈ 0.9184 s

Therefore, both pennies will reach the bottom of the well at approximately the same time, which is 0.9184 seconds. So, there would be no time between the splashes as they would occur simultaneously.

To determine the time between the splash of Terry's penny and Sam's penny, we need to find the time it takes for each penny to reach the bottom of the well.

First, let's find the time it takes for Terry's penny to reach the bottom of the well.

We can use the formula for the time of flight in free-fall motion:

t = sqrt(2h / g),

where t is the time, h is the height, and g is the acceleration due to gravity.

In this case, the height (h) of the well is 29.0 m, and the acceleration due to gravity (g) is approximately 9.8 m/s^2.

Plugging the values into the formula:

t = sqrt(2 * 29.0 / 9.8)
t = sqrt(58.0 / 9.8)
t ≈ sqrt(5.9184)
t ≈ 2.43 seconds

So it takes approximately 2.43 seconds for Terry's penny to reach the bottom of the well.

Now let's find the time it takes for Sam's penny to reach the bottom of the well. Since Sam throws the penny up in the air, it will go up and then come back down. The time it takes for the penny to reach the highest point will be the same as the time it takes for the penny to come back down to the bottom of the well.

Using the same formula:

t = sqrt(2h / g)

With h = 29.0 m and g = 9.8 m/s^2:

t = sqrt(2 * 29.0 / 9.8)
t ≈ 2.43 seconds

So it also takes approximately 2.43 seconds for Sam's penny to reach the bottom of the well.

Since both pennies take the same time to reach the bottom of the well, there is no time between their splashes. They will hit the water simultaneously.