Question 1119939: An object is dropped from the top of ABC Tower, 1,500 feet from the ground. Calculate the average velocity of the object over the interval t=2 seconds and t=3 seconds. The position function of the projectile is s(t) = -16t2 + V0t + h0, where V0t is the initial velocity of the object and h0 is the height to which it is dropped.
Answers:
at t=2 seconds, s(t) = _____ feet
at t=3 seconds, s(t) = _____ feet
average velocity = _____ ft/sec
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! An object is dropped from the top of ABC Tower, 1,500 feet from the ground. Calculate the average velocity of the object over the interval t=2 seconds and t=3 seconds. The position function of the projectile is s(t) = -16t2 + V0t + h0, where V0t [sic] is the initial velocity of the object and h0 is the height to which it is dropped.
Vo is the initial velocity, not Vo*t
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Answers:
at t=2 seconds, s(t) = _____ feet
It's s(2)
Sub 2 for t.
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at t=3 seconds, s(t) = _____ feet
Same as above, t = 3
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average velocity = _____ ft/sec
= s(3) - s(2)
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