Calculating Time in Air for a Vertically Thrown Ball

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A ball thrown vertically upward at 96 ft/s from an 80 ft tall building reaches a maximum height in 6 seconds before falling back to the roof level. The total time in the air is calculated by considering the additional 80 ft it falls after returning to the original height, which takes approximately 0.74 seconds. The correct total time in air is 6.74 seconds, as the displacement equation accounts for the downward motion. The initial velocity remains positive at 96 ft/s for the entire motion, despite the ball moving downward afterward. Understanding the distinction between displacement and total distance traveled clarifies the time calculations.
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Homework Statement


A ball is thrown vertically upward at 96 ft/s from the roof edge of an 80 ft tall building. What is the ball's total time in air (from when thrown).

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The Attempt at a Solution


I calculated that [/B]the ball reaches a height of 0 ft in 6 s. Then to fall -80 ft, I used -80=0+(-96)t-16t^2. I got -96 ft/s as vo in this part of the problem because i assumed the velocity when ball comes back down at the roof's edge is equal and opposite to ball's initial velocity of 96 ft/s. The answer says 6.74 s. But i got 6.74 s to fall that 80 ft. And then I added 6.74 to 6 s to get time in air.
 
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The shape of the ball's trajectory is a parabola. The ball initially moves upwards then it comes back down. When the ball has fallen back to the level that it was thrown at, Δy=0. If we put this into the equation, then you will get 6 seconds, like you calculated. However, the ball keeps moving and falls an additional 80 feet. Δy= -80 feet. The total time it took the ball to move upwards then fall down 80 feet is 6.74 seconds.
 
How'd you get 6.74 s? I'm not following
 
Δy=V0t+1/2at2
Using displacement and not the total distance traveled,
-80 feet= (96 feet/s)t+(1/2)(-32.17 feet/s2)t2
Solve for t, t= -0.74 s and t=6.7 s. A negative time isn't possible so t=6.7s is the right answer.
 
OHHHHHHHH. I understand it now. The delta y applies to displacement and not distance. See but why is v0 positive 96 ft/s. Won't the ball be heading towards and thus in the negative direction?
 
heading downwards is what i meant
 
V0 is the initial velocity for the entire movement. Although the ball's velocity will be -96 when Δy=0, it's not the velocity that is needed for the equation.

Just to be clear, it takes 6 seconds for the ball to reach max height and fall back down to the level it was thrown at. Then it takes 0.74 seconds for it to fall the rest of the way.
 
awesome. thanks a lot.
 
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