Falling object of volleyball

[MIA6]

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?

 

[bel]

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.

 

[m2003]

I would like to clarify that the displacement of -2.0m in this problem represents the initial height of the volleyball above the floor. This value is important because it helps us determine the total distance traveled by the ball, which includes the initial height and the height it reaches at the peak of its trajectory.

To solve this problem, we can use the equation vf^2=vi^2+2ay, where vf is the final velocity (which is 0 when the ball hits the floor), vi is the initial velocity (6.0 m/s upwards), a is the acceleration due to gravity (-9.81 m/s^2), and y is the displacement (which is -2.0m in this case).

Plugging in the values, we get 0^2 = (6.0 m/s)^2 + 2(-9.81 m/s^2)(-2.0 m), which simplifies to 0 = 36 m^2/s^2 + 39.24 m^2/s^2. Solving for t (time), we get t = √(36 m^2/s^2 + 39.24 m^2/s^2)/9.81 m/s^2 = 0.862 seconds.

Therefore, the volleyball will be in the air for approximately 0.862 seconds before it strikes the floor.