KJ4UTS
  • KJ4UTS
Suppose a rock is thrown upwards from the top of a building 200 feet high with an initial velocity of 96 feet per second. The height (in feet) of the rock above the ground t seconds after being released is given by the formula h(t) = -16t^2 + 96t + 200 1.When will it hit the ground? 2.How high will it be after 2.1 seconds? 3.What is the maximum height?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
1. It will hit the ground when h(t) = 0 so you need to solve \[-16t^2+96t+200=0\]
anonymous
  • anonymous
try the quadratic formula
KJ4UTS
  • KJ4UTS
Would it be t=3-√ 43/2=-1.63 and 3+√ 43/2=7.63

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More answers

anonymous
  • anonymous
yes, and since time can't be negative, it hits the ground at 7.63 seconds.
anonymous
  • anonymous
For 2, plug in 2.1 for t to find h(2.1).
KJ4UTS
  • KJ4UTS
331.04
anonymous
  • anonymous
yes. and then for 3 there are two steps. Use \(t=-\frac{ b }{ 2a }\) to find the time of max height. Then plug that time into the h(t) equation
KJ4UTS
  • KJ4UTS
Im a little confused as to what I plug into b and a?
anonymous
  • anonymous
the same values you used in the quadratic equation a = -16, b = 96, c = 200
anonymous
  • anonymous
\[t=-\frac{ 96 }{ 2(-16) }\]
KJ4UTS
  • KJ4UTS
3
anonymous
  • anonymous
now plug 3 in for t in the original equation
KJ4UTS
  • KJ4UTS
Ok I got 344 thank you for explaining this step by step :)
anonymous
  • anonymous
yeah that's it. you're welcome. :)

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