Calculating Time in Air for a Vertically Thrown Ball

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In summary, the conversation discussed the calculation of the total time in air for a ball thrown vertically upward from the roof edge of an 80 ft tall building at a velocity of 96 ft/s. The equation used was -80=0+(-96)t-16t^2, with the assumption that the ball's initial velocity when it comes back down is equal and opposite to its initial velocity of 96 ft/s. The answer obtained was 6.74 seconds, taking into account the additional time it takes for the ball to fall the remaining 80 feet after reaching its maximum height. The conversation also clarified that while the ball's velocity will be -96 when it reaches its maximum height, the initial velocity for the entire movement is
  • #1
Entr0py
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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).

Homework Equations

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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  • #2
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.
 
  • #3
How'd you get 6.74 s? I'm not following
 
  • #4
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.
 
  • #5
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?
 
  • #6
heading downwards is what i meant
 
  • #7
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.
 
  • #8
awesome. thanks a lot.
 

1. What is the definition of ball's time in air?

The ball's time in air refers to the duration that a ball is in the air after it is launched or thrown.

2. How is the ball's time in air calculated?

The ball's time in air can be calculated using the formula t = √(2h/g), where t is the time in seconds, h is the height of the ball in meters, and g is the acceleration due to gravity (9.8 m/s²).

3. Does the ball's time in air depend on its weight or size?

No, the ball's time in air is not affected by its weight or size. It is solely determined by the height from which it was launched and the acceleration due to gravity.

4. How does air resistance affect the ball's time in air?

Air resistance can slightly decrease the ball's time in air by slowing down its velocity. However, this effect is usually negligible for objects like balls that have a small surface area.

5. Can the ball's time in air be longer than its hang time?

Yes, the ball's time in air can be longer than its hang time. Hang time only measures the time between when the ball leaves the hand and when it hits the ground, whereas the ball's time in air also includes the time it takes to reach the maximum height.

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