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At time t = 0, a projectile was fired upward from an initial
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15 May 2019, 05:25
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At time t = 0, a projectile was fired upward from an initial height of 10 feet. Its height after t seconds is given by the function \(h(t) = p−10(q −t)^2\) , where p and q are positive constants. If the projectile reached a maximum height of 100 feet when t = 3, then what was the height, in feet, of the projectile when t = 4 ?
Re: At time t = 0, a projectile was fired upward from an initial
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19 May 2019, 04:45
At t=3, h(3) = p-10(q-3)^2. We are given that height is 100 when t=3 100 = p-10(q-3)^2 100 = p – 10(q^2-6q+9) 100 = p - 10q^2+60q -90--------EQ1 At t =0, we are given that the height is 10 10 = p - 10q^2----Eq2 (EQ1)-(EQ2) 90 = 60q-90 180 = 60Q q= 3 Lets substititute q in EQ2 10-p = -90 -p=-100 P = 100 H(4) = 100-10(3-4)^2 H(4) = 100-10 = 90
Re: At time t = 0, a projectile was fired upward from an initial
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06 Dec 2024, 04:42
Hello from the GRE Prep Club BumpBot!
Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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