A science student drops a rok down a mine shaft. It takes 3 sec. for the rock to hit the bottom of the shaft , at what speed of the rock just before impact?
32 feet per second per second is the acceleration rate of earth. At the end of second 1 the speed is 32 feet per second, at the end of second 2 it is 64f.p.s., At the and of second 3, 96f.p.s. minus atmospheric drag. Ask your teacher about Terminal Velocity.
Well, let me tell you, that rock must have been feeling pretty down during that drop! As for the speed just before impact, we can use a little bit of math (but I promise, it won't be rock-et science).
Since the acceleration rate of free fall on Earth is 32 feet per second per second, we can multiply it by the time it took for the rock to drop, which is 3 seconds. That gives us a speed of 96 feet per second, but remember to subtract the effect of atmospheric drag if you want to be more accurate.
So, the rock was cruising at a speed of 96 feet per second just before it met its rocky fate. I hope it didn't land too hard, or else it might have gotten a bit "sedimentary" down there!
To calculate the speed of the rock just before impact, we can use the formula for calculating velocity based on acceleration and time:
velocity = acceleration x time
In this case, the acceleration is due to gravity and is equal to the acceleration rate of Earth, which is 32 feet per second squared (32 ft/s^2). The time it takes for the rock to hit the bottom of the shaft is 3 seconds.
Using the formula, we can calculate the velocity:
velocity = 32 ft/s^2 x 3 s
velocity = 96 ft/s
Therefore, the speed of the rock just before impact is 96 feet per second. However, it is worth noting that this calculation assumes there is no atmospheric drag affecting the speed of the rock. In realistic scenarios, objects may experience a terminal velocity due to the opposing force of the air resistance. To accurately account for atmospheric drag, it is advisable to consult a teacher or reference material for further clarification.