Sam and Terry find a 21.0 m deep wishing well and decide to make a wish. Terry throws a penny down the well at 7.00 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?
The depth of the well does not matter.
How long does it take the coin thrown upwards to come back down to the top of the well?
7.00 - 9.81t = 0
That is how long it takes to stop rising and start falling. Double that, and you have the time between splashes.
Well, well, well! Let's do some penny-pitching physics, shall we? Since Sam and Terry threw their pennies at the same speed, and we're not accounting for any air resistance, we can focus on the upward and downward motion separately.
Let's consider Terry's penny first. He threw it downwards at 7.00 m/s and the well is 21.0 m deep. To find the time it takes for Terry's penny to splash, we can use the equation:
t = d/v
Where:
t = time
d = distance
v = velocity
Plugging in the values, we get:
t = 21.0 m / 7.00 m/s
t = 3.00 seconds
So, it would take Terry's penny 3.00 seconds to reach the bottom of the well.
Now, onto Sam's penny. Since Sam threw it upwards at the same speed, the penny will reach its highest point at the same time as Terry's penny splashes. So, the time between splashes would be the same as the time it takes for Sam's penny to reach its highest point and then come back down.
In this case, we need to consider the height of the well as the total distance traveled by Sam's penny (21.0 m up + 21.0 m down). Using the same equation:
t = d/v
t = (21.0 m + 21.0 m) / 7.00 m/s
t = 6.00 seconds
Therefore, the time between splashes would be 6.00 seconds.
So, the answer is that there would be a "well" balanced 6.00 seconds between the two splashes. Enjoy your physics-themed wish-making session!
To determine the time between the splashes, we can use the equation of motion for each penny.
For Terry's penny going down, we have:
Initial velocity (u) = 7.00 m/s (downward)
Acceleration (a) = acceleration due to gravity = -9.8 m/s^2 (since it's acting in the opposite direction)
Displacement (s) = -21.0 m (negative because it is moving downward)
Using the equation of motion:
s = ut + (1/2)at^2
Rearranging the equation, we get:
t^2 + (2u/a)t + (2s/a) = 0
Substituting the values, we have:
t^2 + (2(7.00)/(-9.8))t + (2(-21.0)/(-9.8)) = 0
Simplifying the equation, we have:
t^2 - 1.4286t + 4.2857 = 0
Next, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, we have:
a = 1
b = -1.4286
c = 4.2857
Substituting the values, we have:
t = (-(-1.4286) ± √((-1.4286)^2 - 4(1)(4.2857))) / (2(1))
Simplifying further, we get:
t = (1.4286 ± √(2.0408 - 17.1428)) / 2
Since the square root term results in a negative value, it means that the penny will not hit the bottom of the well. Therefore, we discard that negative solution.
So, the time it takes for Terry's penny to hit the bottom of the well is approximately:
t = (1.4286 + √(-15.102)) / 2
≈ (1.43 + √(-15.102)) / 2
Now, let's calculate the time it would take for Sam's penny to reach its highest point.
For Sam's penny going up, the final velocity at the highest point is 0 m/s since it changes direction.
Using the equation of motion:
v = u + at
Rearranging and substituting the values, we have:
0 = 7.00 - 9.8t
Simplifying the equation, we get:
9.8t = 7.00
t = 7.00 / 9.8
t ≈ 0.714 seconds
Since Sam's penny takes 0.714 seconds to reach its highest point, we can assume that it will also take the same time to come back down and reach the same height as Terry's penny.
Therefore, the total time between the splashes would be approximately:
2 * 0.714 seconds = 1.428 seconds
So, there will be roughly 1.428 seconds between the splashes.