Throwing a Ball Horizontally from a Building: Physics Analysis

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A ball thrown horizontally from a building lands a distance d after time T, assuming negligible air resistance. If the building's height is doubled, the time in the air becomes T squared root 2, leading to the conclusion that the ball would land at a distance of 2d. The horizontal velocity remains constant, meaning the ball would also reach the ground with twice the speed if it were in the air for time 2T. The equations of motion confirm these relationships, highlighting the interdependence of distance, time, and velocity in projectile motion. Understanding these principles is crucial for solving similar physics problems.
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A ball is thrown horizontally from the top of a building and lands a distance d from the foot of the building after having been in the air for a time T and encountering no significant air resistance.

If the building were twice as tall, the ball would have

a) landed a distance 2d from the foot of the building
b) been in the air a time 2T
c) been in the air a time T squareroot2
d) reached the ground with twice the speed it did from the shorter building


x=x0 + v0xt + 1/2at^2
vx = v0xt +axt
and vx^2 = V0x^2 +2a (xf - xi)

Are, I think, the right equations. I don't know what to do though :(
 
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D'oh, it was C. Thanks anyways :)
 
Note that the other three are equivalent: if the ball is in the air for a time 2T and the horizontal velocity does not change (no resistance) it will land at a distance 2d and since the velocity is v = a T it would also go twice as fast. So if any of these were true, then the other three would be true as well. Only b and c are clearly contradictory, so you just had to decide which one of them was true :smile:
 
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