I came across a question and it asked for the time it took for an object with an initial velocity to go up and come down. I was just wondering why the time it takes to go up for the object is the same as the time it takes for the object to fall to it's original location. If an object going up had a non-zero initial velocity (11), however going down it's initial velocity is 0, given that the acceleration and distance is the same, wouldn't the time taken for going up and going down be different.
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2 Answers
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It's a symmetric situation. Instead of looking at INITIAL velocity, recognize that the upward motion and downward motion are, point-for-point, mirror images. The average upward velocity in the rise, and average downward velocity in the fall, match in magnitude (and are opposite in sign). The displacement in the rise and in the fall are matched in magnitude (and opposite in sign). So, the times are the same.
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While going up, it has initial velocity x and final velocity zero, with acceleration -g.
When it is coming down, initial velocity is zero and distance is same as when going up (opposite direction). Acceleration is +g. You can see from the equations that time turns out to be same for both cases. And thr final velocity is x too (energy conservation).
