-1
$\begingroup$

Identical balls & building , assuming vacuum & ignoring terminal v altogether, one dropped and one thrown directly down at same moment, how could one theoretically achieve a simultaneous landing?

$\endgroup$
2
  • $\begingroup$ I am on a bike, a professional cyclist leaves the start gate (even at the same time as me), identical bike etc, am I going to catch him, no.... $\endgroup$
    – user140606
    Jan 7, 2017 at 15:49
  • $\begingroup$ -1. No effort to tackle the question. $\endgroup$ Jan 7, 2017 at 19:43

1 Answer 1

2
$\begingroup$

You can't. If you're imparting additional force to one of them and release both at the same point, the thrown object will will land first.

Also, if you assume you're in a vacuum, you don't need to ignore terminal velocity. There is none.

EDIT: I suppose you could do it if the ground weren't completely level, but I assume you mean all things are equal.

$\endgroup$
4
  • $\begingroup$ +1 for being 4 secs ahead of me, I will never catch up... $\endgroup$
    – user140606
    Jan 7, 2017 at 15:51
  • $\begingroup$ Yes vacuum, would there not be a way in theory to accelerate outside of g $\endgroup$
    – ARinLA
    Jan 7, 2017 at 16:16
  • $\begingroup$ If balls are identical but one could could have additional force acting on it prior to the other - maybe it the velocity of the thrown ball is so high that it approaches c .. thus time slows down $\endgroup$
    – ARinLA
    Jan 7, 2017 at 16:21
  • $\begingroup$ @ARinLA That isn't how relativity works. To an observer in this situation, the only differences would be that a fast-moving ball would appear to be rotated toward the observer as it passes them, a clock on that ball would move more slowly, and it would take more and more acceleration to increase velocity by a marginal amount. It wouldn't change the relative velocity of the ball. $\endgroup$ Jan 7, 2017 at 16:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.