You drop a rock into a well and count the seconds before you hear a splash. If you counted 10 seconds before you heard the splash, how deep is the well?
if the speed of sound is x m/s, then if the well's depth is d,
4.9(10-d/x)^2 + d/x = 10
now look up x and solve for d
Have to differ on this one Steve. You have two distances adding up to 10? I doubt they want any more serious analysis than 1/2 g t^2 (although 490m is pretty deep and with the speed of sound at 343 m/s it IS significant).
But let's go your way.
Time down as a function of dist
t1 = sqrt(d/4.9)
Time back
t2 = d/343
t1+t2 = 10
Solving this (with much difficulty) yields the quadratic
d^2-30870d+11764900 = 0
And the well is about 386 deep.
So I'm guessing they want the simple solution.
To determine the depth of the well, we can use the principles of free fall and the constant acceleration due to gravity. The formula to calculate the distance fallen is:
d = (1/2) * g * t^2
Where:
d = distance fallen (depth of the well)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken (10 seconds in this case)
Plugging in the given values into the equation:
d = (1/2) * 9.8 m/s^2 * (10 s)^2
Simplifying the equation:
d = 0.5 * 9.8 m/s^2 * 100 s^2
d = 4.9 * 100 m
Therefore, the depth of the well is approximately 490 meters.
To determine the depth of the well, you can use the equation of motion for a falling object. The equation is:
d = (1/2) * g * t^2
Where:
- d is the depth of the well
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken for the rock to fall
In this case, we have a time of 10 seconds. So, we can substitute the values into the equation:
d = (1/2) * 9.8 * (10)^2
d = 0.5 * 9.8 * 100
d = 490 meters
Therefore, the depth of the well is approximately 490 meters.