If a girl drops a coin into a wishing well. After exactly 5 seconds she hears it splash into the water. How fast was the coin going after 1 second? After 2 seconds? How fast was the coin going when it hit the water? How far down is the water? Could you please help with the formulas and the answers for these questions?
OK, here is how to do that:
How fast was the coin going after 1 second? V(1) = 1*t = 9.8 m/s
After 2 seconds? V(2) = 2*t = 19.6 m/s
How fast was the coin going when it hit the water? V(T) = 5*T
Let T be the time when it hit the water.
It will also take the sound of the splash 340*T to be heard. 340 m/s is the assumed speed of sound.
How far down is the water?
Let H be the depth of the water in the well.
sqrt(2H/g) + H/340 = 5
You will have to solve that graphically, by iteration or the quadratic formula
H = 108 meters
Time when it hit water:
T = sqrt(2H/g) = 4.69 s
V(4.69) = 46 m/s
To solve this problem, we'll need to use the equations of motion and assume that there is no air resistance. Let's break down the problem step by step:
1. How fast was the coin going after 1 second?
To determine the velocity of the coin after 1 second, we can use the equation of motion:
v = u + at
Here:
v = final velocity (unknown)
u = initial velocity (unknown)
a = acceleration (which will be the acceleration due to gravity, approximately -9.8 m/s^2)
t = time (1 second)
Since the coin is dropped, its initial velocity (u) is 0 m/s (as there is no initial push).
Plugging in the values, the equation becomes:
v = 0 + (-9.8) * 1
v = -9.8 m/s
Therefore, after 1 second, the coin is traveling at a speed of -9.8 m/s. The negative sign indicates that the velocity is directed downwards.
2. How fast was the coin going after 2 seconds?
Using the same equation and reasoning, we can find the velocity of the coin after 2 seconds:
v = u + at
Here:
v = final velocity (unknown)
u = initial velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = 2 seconds
Plugging in the values, the equation becomes:
v = 0 + (-9.8) * 2
v = -19.6 m/s
Therefore, after 2 seconds, the coin is traveling at a speed of -19.6 m/s.
3. How fast was the coin going when it hit the water?
Since the coin hits the water after 5 seconds, we can use the same equation to find the velocity at that moment:
v = u + at
Here:
v = final velocity (unknown)
u = initial velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = 5 seconds
Plugging in the values, the equation becomes:
v = 0 + (-9.8) * 5
v = -49 m/s
Therefore, when the coin hits the water, it is traveling at a speed of -49 m/s.
4. How far down is the water?
To calculate the distance the coin has fallen (i.e., the depth of the well), we can use the formula:
s = ut + 0.5 * a * t^2
Here:
s = distance (unknown)
u = initial velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (5 seconds)
Plugging in the values, the equation becomes:
s = 0 + 0.5 * (-9.8) * (5^2)
s = 0 + 0.5 * (-9.8) * 25
s = 0 + (-122.5)
s = -122.5 meters
Therefore, the water is 122.5 meters down the well.
Note: The negative sign indicates that the distance is measured downward in the opposite direction of the coin's motion.
So, to summarize:
- After 1 second, the coin is traveling at -9.8 m/s.
- After 2 seconds, the coin is traveling at -19.6 m/s.
- When the coin hits the water after 5 seconds, its velocity is -49 m/s.
- The depth of the water in the well is approximately 122.5 meters.