SOLUTION: Help please!
H=vt +1/2at^2
H=25,000
t=10
v=2,000
Does the acceleration exceed 3g(1g=32ft/sec^2)?
Thank you so much.
Algebra.Com
Question 733739: Help please!
H=vt +1/2at^2
H=25,000
t=10
v=2,000
Does the acceleration exceed 3g(1g=32ft/sec^2)?
Thank you so much.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
H=vt +1/2at^2
H=25,000
t=10
v=2,000
Does the acceleration exceed 3g(1g=32ft/sec^2)?
--------------
H=vt + at^2/2
25000 = 2000*10 + 50a
50a = 5000
a = 100 ft/sec/sec, > 96
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