During a baseball game, a batter hits a

pop-up to a fielder 82 m away.
The acceleration of gravity is 9.8 m/s2 .
If the ball remains in the air for 6.9 s, how
high does it rise?
Answer in units of m.

5 answers

goes up for 6.9/2 = 3.45 seconds
initial speed up = Vi
v = Vi - 9.8 t
at top, v = 0
0 = Vi - 9.8 (3.45)
Vi = 33.8 m/s

h = Vi t - 4.9 t^2
= 33.8(3.45) - 4.9(3.45^2)
= 116.6 - 58.3
=58.3 meters
note, the horizontal proble, the 82 m range, has nothing to do with the problem so far.
THANK YOU
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Its been ten years my guy I don't think they are gonna respond
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