Known: v, t, and a cannot find Δy

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The discussion revolves around solving a physics problem involving a basketball thrown straight down and its subsequent bounce. The poster struggles to find the ball's maximum height after it bounces back up, given the total time of 2.5 seconds for the entire motion. Initial calculations using the equations of motion yield incorrect results, leading to confusion about the need for initial velocity. Suggestions from others emphasize analyzing the complete motion and correctly applying the equations, particularly focusing on the time split during ascent and descent. Ultimately, the poster acknowledges the need to double-check calculations to arrive at the correct answer.
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Known: v, t, and a ... cannot find Δy

Long time lurker, first time poster. Thanks for all the help y'all provide.

Problem #1:

Homework Statement



Wrongly called for a foul, an angry basketball player throws the ball straight down into the floor. If the ball bounces straight up and returns to the floor 2.5s after first striking it, what was the ball's greatest height above the floor?

Homework Equations



Only know three so far:

Y=Y_0 + Y_0t + ½ at²
V=V_0 + at
V²=V_0² + 2aΔx

The Attempt at a Solution



Used the top equation, since I know t and need to know y. I'm under the impression that this is a freely-falling objects question, so:

y_0 = 0 (the floor)
initial velocity = 0, at the very point the ball reverses direction to start its ascent
t = 2.5 seconds (given)
a = -9.8 m/s² (gravity)

y = 0 + 0t + ½(-9.8)(2.5)² = 30.625m

This is an incorrect answer, of course. I keep thinking there needs to be an initial velocity, but it's not given, and the only other way I know to calculate velocity is Δx/Δt, and I'm obviously trying to find Δx (or rather, Δy in this case). Any pointers would be most appreciated. Thanks!
 
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If you use your equation to analyze the complete motion of the ball as it goes from ground to ground, you'll be able to solve for the initial speed. That's a good first step.
 


Doc Al said:
If you use your equation to analyze the complete motion of the ball as it goes from ground to ground, you'll be able to solve for the initial speed. That's a good first step.

OK, conceptually I get that it's a parabola, and there's a negative portion to the parabola, but wouldn't I need to know at least one other factor to figure that out? If I had the initial velocity as the ball leaves the player's hands, I'd know v and a and could figure out Δy and then t, right? If I knew the distance traveled from the player's hands to the floor, I could figure out the initial velocity, right?

I know this is supposed to be easy, but I'm just missing something obvious here.
 


Hold on, I think I have it ...
 


OK, I don't have it, but here's what I tried:

From the apex of the ball's flight to the ground, the initial velocity is zero (at apex). Using v = v_0 + at and halving the time, I get ...

v = 0 - 9.8(1.25) = -24.5 m/s ; this should be the velocity immediately before the ball hits the floor on its descent.

When I use that velocity as the initial velocity in the first part, it comes out to ~92m, which is incorrect. So I decided to try the equation in reverse. If the initial velocity is 24.5, the time is 1.25 (half the total time?), and the acceleration is -9.8, I should be able to get the total height, right? Not so much, it seems.

y = 0 + 24.5(1.25) - 4.9(1.25)^2 = 22.97m

That's not correct, either.
 


eriadoc said:
OK, I don't have it, but here's what I tried:

From the apex of the ball's flight to the ground, the initial velocity is zero (at apex). Using v = v_0 + at and halving the time, I get ...

v = 0 - 9.8(1.25) = -24.5 m/s ; this should be the velocity immediately before the ball hits the floor on its descent.
Double check your arithmetic.
 


Doc Al said:
Double check your arithmetic.

D'oh! I was using the 2.5 number for time in that part of the calc.

Sir, you are a scholar and a gentleman. Thank you immensely.
 
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