Baseball: Rising and Falling speed

AI Thread Summary
The discussion revolves around a physics problem involving the motion of a baseball thrown upward. The initial speed of the baseball is calculated using the formula Vf^2 = Vi^2 + 2ad, yielding an initial speed of 22 m/s. For part (c), the time taken to pass the window is found to be approximately 1.4 seconds, based on the initial speed. There is uncertainty regarding the correct acceleration value to use when calculating the upward motion versus free fall. The participant seeks assistance with parts (b) and (d) of the problem, specifically regarding the maximum altitude and the time taken to return to the street.
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Homework Statement



A baseball is seen to pass upward by a window 28 m above the street with a vertical speed of 8 m/s. The ball was thrown from the street.

(a) What was its initial speed?
(b) What altitude does it reach?
(c) How long after it was thrown did it pass the window?
(d) After how many more seconds does it reach the street again?

Homework Equations



Vf = Vi + at
Vf^2 = Vi^2 + 2ad
d = Vi(t) + (.5)(a)(t)^2
t = Vf/a or Vf-Vi/a



The Attempt at a Solution



So far I've only attempted a and c because I don't know how to go aout solving b or d.

For a, I used the Vf^2 formula and changed it to equal Vi^2. So I had:

2(a)(d) - Vf^2 = Vi^2 and with that I got 2(9.8)(28) - (8)^2 = Vi^2 and got 22

I believe the problem may be with acceleration. Is it 9.8 when thrown uo, or only during free fall?


For c, I took the answer i got from a, Vi = 22, and plugged it into the time equation. I did 8 - 22 / 9.8 to get 1.4 seconds.
 
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I'm not sure if my answers for a and c are correct. If someone could help me with b and d, I'd really appreciate it.
 
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