Find Max Height for Basketball Throw: Is 9.8 Correct?

AI Thread Summary
The discussion centers on calculating the maximum height a basketball reaches after being thrown down and bouncing back up. A player throws the ball down, and it takes 2.8 seconds to return to the floor, indicating that the time to reach maximum height is 1.4 seconds. Participants point out that the original calculations are overly complicated and suggest using the simpler formula y = 1/2 * g * t^2 to find the height. The correct approach involves understanding the relationship between time, gravity, and height. Ultimately, the focus is on simplifying the calculation to accurately determine the ball's greatest height above the floor.
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Wrongly called for a foul, an angry basketball player throws the ball straight down to the floor if the ball bounces straight up and returns to the floor 2.8 s after first striking it, what was the ball's greatest height above the floor?

I keep getting the wrong answer I am plugging in (9.8)(1.4)(.5)(1.4)^2(9.81)

Am I anywhere close?
 
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wadini said:
Wrongly called for a foul, an angry basketball player throws the ball straight down to the floor if the ball bounces straight up and returns to the floor 2.8 s after first striking it, what was the ball's greatest height above the floor?

I keep getting the wrong answer I am plugging in (9.8)(1.4)(.5)(1.4)^2(9.81)

Am I anywhere close?

Welcome to PF.

Well ... looks like a few too many factors to have the right answer.

You know the time and so 1/2 will be time to max height ... you got that part.

But the distance it will fall in 1.4s is really given a bit more simply by

y = 1/2*g*t2
 
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