EugP said:
Think of it this way. What is acceleration? It's change of velocity over time. Well if you have a short period of time, like if you drop a penny one inch above your hand, you won't have a big change in velocity. But if you drop it 3000 feet above your hand, it will take it longer, meaning there will be more change in velocity. In other words, if it doesn't have to travel a large distance, it won't have time to gain a lot of speed.
I hope this helped.
But for an object falling, doesn't its acceleration remain constant throughout the entire fall?
For the penny falling a short distance, if inital velocity = 0, and final velocity = 10, and its acceleration is 9.81 m/s/s, then the time it takes to fall will be 0.98 s.
For the penny falling a long distance, inital velocity = 0 and final velocity = 50 (hypothetically only, just as long as it's bigger than 10, because it has more time to get up to a velocity that high). Its acceleration, however, remains the same at 9.81 m/s/s - it just takes longer to fall: 5.1 s this time.
Right? There is a bigger change in velocity, but because the time interval is also bigger, the acceleration remains the same.
I guess my question now is, why does speed affect the fact that the penny hits harder when dropped from a higher distance, since the formula f=ma doesn't account for speed?
mgb_phys said:
Now the time taken for the penny to be stopped by your hand is roughly the same, the end velocity is zero (the penny is stopped. What is different about the velocity of the two pennies when they are just about to touch your hand ?
The long distance penny has a higher velocity, so the deceleration of this same penny is higher... using f=ma also means that that force is bigger. I think I kind of get it now!
So why isn't it the accceleration of the falling penny that is used to calculate this, but its deceleration after it hits my hand?
