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Message: <blockquote data-quote="Dausuul" data-source="post: 4102895" data-attributes="member: 58197">Air resistance is the only thing that prevents Broghammerj from being correct.  In a vacuum, he would be exactly right.  If we fire a bullet at an upward angle, at a target at the same altitude:The lateral component of its velocity remains unchanged thanks to Newton's First Law, at least until it crashes into something.The vertical component of its velocity changes at a constant rate.  It starts out going upward with a speed of X meters/second (for some value of X depending on the bullet, the gun, et cetera).  Gravity accelerates it downward at 9.8 m/s/s.  So after X/9.8 seconds, it will have a vertical velocity component of zero--it is neither rising nor falling.The formula for the bullet's upward speed is X - 9.8t meters/second.  Integrating that over t will give us the height at time t, which is Xt - 4.9t^2.  Plug in X/9.8 for t, and we end up with (X^2)/19.6.  That's the height at which the bullet stops rising and starts falling.After that, it starts going downward.  As you say, physics has no memory, so (disregarding the lateral component) this is exactly as if we'd simply dropped the bullet from a height of (X^2)/19.6 meters.  We'll reset the clock to zero in order to simplify the math.Gravity is still accelerating it at 9.8 m/s/s, so at time t, the bullet's downward speed is 9.8t meters/second.  Again integrating over t to get the distance fallen, we get 4.9t^2.The bullet hits the ground when the distance fallen equals the height from which it was dropped, which is to say that 4.9t^2 = (X^2)/19.6.  Solve for t, and you get t = X/9.8 seconds.  Plug that into the formula for speed, and the downward speed is X--exactly the same as the upward speed was when we originally fired it.</blockquote>

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