You are standing on top of a building. Two balls are thrown

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The discussion centers on the physics of two balls thrown from a building: one upwards at 10 m/s and the other downwards at the same speed. Despite the upward ball gaining gravitational potential energy, the downward ball is expected to reach the ground first due to its initial downward velocity and the acceleration of gravity. Both balls experience gravitational acceleration at 9.8 m/s², but the upward ball must first decelerate to a stop before falling. The key takeaway is that the ball thrown downwards will arrive at the ground before the one thrown upwards, despite the latter having greater kinetic energy upon impact. Understanding these principles illustrates the effects of initial velocity and gravitational acceleration on falling objects.
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One ball is thrown upwards at 10 m/s, the other is thrown straight down (not dropped) also at 10 m/s. Which ball arrives first?

This is more of a theoretical question than anything. I know the relevant equations such as v^2=u^2 +2as, but I was thinking that the height of the building has to count for something because the ball traveling straight upwards would gain gravitational potential energy. But "common sense" dictates that the ball thrown straight down would arrive at the floor first.
 
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Remember that everything accelerates (generally) at 9.8 m/s^2 towards the ground due to gravity. While the higher ball thrown up will have a greater gravitational potential it will also have a higher kinetic energy when it hits the ground, this is because it has been falling a greater distance. That does not mean that it will manage to accelerate any faster than the ball thrown downwards.
 
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