Sam and Terry find a 30.0 m deep wishing well and decide to make a wish. Terry throws a penny down the well at 3.00 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?
how long does it take a penny to rise and fall, starting at 3.0 m/s?
v = 3 - 9.8t
double that, and that's how long the first penny has been dropping before the 2nd gets back to the top of the well.
To find the time between the splashes, we first need to determine the time it takes for each penny to reach the bottom of the well.
Let's start with Terry's penny:
We can use the equation of motion to find the time it takes for Terry's penny to fall the 30.0 m depth of the well.
The equation of motion for an object in free fall is:
d = vi * t + (1/2) * a * t^2
Where:
d is the distance traveled (30.0 m),
vi is the initial velocity (3.00 m/s),
t is the time,
a is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the penny is thrown downward, the acceleration will be -9.8 m/s^2.
Plugging in the values, we get:
30.0 m = 3.00 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Rearranging the equation to solve for time, we have a quadratic equation:
-4.9 t^2 + 3.00 t - 30.0 = 0
Solving this equation, we find that the time it takes for Terry's penny to reach the bottom of the well is approximately 2.72 seconds.
Now let's calculate the time it takes for Sam's penny to reach the highest point and come down to the bottom of the well.
Since Sam's penny is thrown up at the same speed of 3.00 m/s, the time to reach the highest point will be the same as the time to fall down to the bottom of the well.
Therefore, the total time for Sam's penny to reach the bottom of the well is also approximately 2.72 seconds.
As a result, the time between the splashes would be the same as the time it takes Terry's penny to reach the bottom of the well, which is approximately 2.72 seconds.