Using classical mechanics, a body is projected upwards from the origin at a velocity v than falls to the ground. Ignoring air resistance, it can be shown that the velocity of the ball on impact at the origin is equal in magnitude to the initial velocity v. A bit of a more realistic model. Allow for air resistance for a sphere of a given mass & diameter, with initial velocity v. When the ball falls back to the origin, will the speed be of equal magnitude to the initial velocity? If I take all my calculations from the origin, I can show it is - though I may have made a dodgy assumption doing it this way). If I go to the max high point, reset that as the origin with initial velocity of zero, I get a different (lower value ) answer (my calc's for h max are correct). This goes against my intuition that both values should be equal, I've checked over and over and I'm pretty confident that the calculations (second method) are correct. So, either my intuition (pre-conceived ideas) are wrong, or there is a mistake in there somewhere. Should I keep checking, or should I accept what the maths is telling me?

I'm no stunning mathematician, but I think your conclusion that the speed of the ball when it returns to the ground will be the same as it was launched at is incorrect. Certainly if the ball is launched at above its terminal velocity it will be incorrect, and I think in all other cases as well. When the ball is launched upwards, it will be imparted with a certain amount of kinetic energy. This is converted to potential energy as the ball approaches its apogee and is converted back into kinetic energy as it falls back to ground. However, if the ball is travelling through air, it is constantly leaching that energy into the air that it pushes out of the way. As for your maths, are you sure you're using exactly the same equations in both cases? It shouldn't make any difference where you start the system from.

Yes, so by the conservation of energy it can be shown that (neglecting air resistance) the speed of ball is the same in both cases is the same since no energy is lost. Taking air resistance into acount, energy is lost - so not all of the K.E. is converted into P.E. Then at H max, some of the P.E. is lost during the fall - and so the magnitude of the speed of the ball at the origin is less than the initial velocity v. And so my calcs are correct (though I'll have to look again at that dodgy assumption). Cheers - you've shown me the light

This for OU or are your preparing for it? How you getting on with it in general? I use Bostock and Chandler applied mathematics v1&2 for mechanics.

O.U. task. But I was so wrapped up in deriving the equation of motion and the integration that I couldn't believe what the maths was telling me. Thinking about it in terms of conservation of energy (thanks again Tim), all of my calculations make sense. All sorted for now.

Sad case that I am, I was wondering about this kind of thing the other night while watching the news... A group of excitable gentlemen in a crowd were firing shots into the air. I wondered how fast the bullets would be going when they came back down! I think a lot of energy would be lost in air resistance because the bullet would initially be travelling extremely quickly. I'd certainly much rather be hit by a falling bullet than by one that just been fired. I did wonder if the falling bullets could kill someone and, Google being my Friend, the answer is... possibly!

To be fair, it's unlikely to come down in exactly the same place. If you're interested in why even on an approximate level look up some topics in fluid dynamics such as magnus effect, yaw causing spin drift, reynolds number, equation of drag and other related matters. There are ballistics tables that tell you all sorts of things. I'm interested in fluid dynamics in general though, rather than ballistics.

Oh, and in answer to the original question, - as PBT said, terminal velocity is the limit. The object is projected upwards with an initial velocity, decelerating under the effect of gravity until the velocity is zero. It then starts accelerating downwards under gravity (9.8ms^-2) until it reaches a limit, either reaching terminal velocity or hitting the ground

its velocity will be greater when launched than when coming back down, even if it doesn't reach terminal velocity.

you can work it out. All you need to know is the force due to gravity is [some constant], the force due to wind resistance is [some constant] x velocity squared, acceleration is [some constant (i.e. mass)] x acceleration, and velocity is acceleration x time. Make the equations from there, and fit them together. i might do it for you some time tomorrow if i can be arsed but i'm not staying up much longer.