How do I calculate distance fropped from time and acceleration?
- Saturday Dec 26,2009 07:28 PM
- By diddy
- In Others
I wanted to know how deep my borehole is approximately so I dropped a stone down the shaft and counted about 3 seconds. It was a straight drop without any hitting the walls of the shaft. How deep is the borehole?
Acceleration, Borehole, Calculate Distance, Shaft, Straight Drop
3 Comments
you can use the accelerated motion, in which a = g, gravity acceleration:
H = h0 + v0t + gt²/2
since the starting height is.. ummm, zero, and you just dropped it down, we’ll have:
H = gt²/2
H = 10*9/2
H = 45m
this is how deep your borehole is: 45m, not considering air resistance;
Use:
distance(height) = 1/2 * acceleration(gravity) * (time)^2
d = 1/2 g t^2
you know g(9.80m/s/s) and t(3 seconds)
Solve for distance, or how deep the hole is.
Don’t forget speed of sound.= 345 m/s (approx) at standard temperature & pressure.
The stone is dropped and falls under "constant acceleration" but the sound of the stone hitting bottom takes some time to travel back to the listener. So to be accurate (assuming no drag of air friction on the stone during its fall) the total time of 3 sec includes the sound travel time.
If t1 = time for stone’s fall, and t2 = time for sound to travel depth (h) of borehole.
3 = t1 + t2
t2 = (3 - t1)
h = 1/2g(t1)²
h = (345)(t2)
1/2g(t1)² = (345)(t2)
(0.5)(9.8)(t1)² = (345)(3 - t1)
4.9(t1)² = 1035 - 345(t1)
4.9(t1)² + 345(t1) - 1035
solve quadratic for pos root of "t1":
t1 = 2.882 s
t2 = 0.118 s
h = (345)(0.118) = 40.7 m ANS
h = 4.9(2.882)² = 40.7 m check ANS
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