SOLUTION: Dear... Can someone help me with this? A projectile is fired straight upward from ground level with an initial velocity of 80 feet per second. During which interval of time

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Question 1074061: Dear...
Can someone help me with this?
A projectile is fired straight upward from ground level with an initial velocity of 80 feet per second.
During which interval of time will the projectile's height exceed 96 feet?
I do not know how to use the position formula here.
Thank you.
Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
the problem uses feet, so we will use 16 feet per second per second which is derived from the force of gravity
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the formula for height at a give time is
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h(t) = -16t^2 + v(0)t + h(0)
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Note that h(0) is 0 at ground level
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the graph of h(t) is a parabola that open downward
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we are given v(0) = 80 feet per second
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h(t) = -16t^2 +80t
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we want to find the vertex of the parabola by taking the first derivative of h(t) and setting it = 0 and solve for t
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h'(t) = -32t + 80
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-32t = -80
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t = 2.5 seconds
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this tells us that the maximum height of the projectile occurs at 2.5 seconds after launch
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h(t) = -16(2.5)^2 + 80(2.5) = 100 feet
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The maximum height is 100 feet at 2.5 seconds after launch
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now we look for the time after launch when the projectile reaches 96 feet
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96 = -16t^2 + 80t
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16t^2 -80t +96 = 0
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divide both sides of = by 16
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t^2 -5t +6 = 0
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factor the quadratic
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(t-3) ^ (t-2) = 0
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t is 3 or 2
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Note the max height occurs at 2.5 seconds
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The projectile's height exceeds 96 feet during the time interval (2, 3)
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here is the graph of the projectiles path
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