How Long Will the Volleyball Be in the Air?

AI Thread Summary
The discussion focuses on calculating the time a volleyball is in the air after being hit straight upward with an initial velocity of 6.0 m/s from a height of 2.0 m. The key equation used is vf^2 = vi^2 + 2ay, where the displacement is considered as -2.0 m, indicating the volleyball's final position relative to its starting point. There is confusion regarding the significance of the displacement value, with clarification that it represents the vertical distance traveled downward to the floor. The trajectory of the volleyball is described as a parabolic curve, which results in two potential time values, necessitating the rejection of one to find the correct duration. Understanding the displacement in this context is crucial for accurately determining the time the volleyball remains airborne.
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Homework Statement


Jason hits a volleyball so that it moves with an initial velocity of 6.0 m/s straight upward. If the volleyball starts from 2.0 m above the floor, how long will it be in the air before it strikes the floor?


Homework Equations


y=-2.0m,Vi=+6.0 m/s a=-9.81m/s^2
vf^2=vi^2+2ay. then just put the numbers in it.

The Attempt at a Solution


I don't understand why the displacement here is -2.0m? I don't think it can be used in solving the problem because 2.0m is not the distance that the ball traveled, but just 2.0m above the floor? What's the significance of it?
 
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The displacement is -2.oo m because that's the distance it ended up at from the starting point. If you think about the displacement time curve, it's a parabola, so if you read across the graph for a displacement of -2.oom, you have two time values, one of them has to be rejected to give the right answer.
 
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