Physics
posted by Briana .
Before working this problem, review Conceptual Example 15 . A pellet gun is fired straight downward from the edge of a cliff that is 22.3 m above the ground. The pellet strikes the ground with a speed of 34.9 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?

h=v₀t +gt²/2
v= v₀+gt
v₀ =vgt
h=( vgt)t + gt²/2 =
=vt  gt² + gt²/2=
= vt  gt²/2.
gt² 2vt +2h = 0
t = (2•34.9±sqrt{(2•34.9)²4•9.8•2•34.9}/2•9.8 = 0.7 s
v₀ =vgt =34.9 – 9.8•0.7 = 28 m/s
Upward motion
H=v₀²/2g= 28²/2•9.8 = 40 m
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