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October 22nd, 2009, 06:19 PM   #1
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Differential Equation: Motion Problem

A canyon is 1100 meters deep at its deepest point. A rock is dropped from the rim above this point. Write the height of the rock as a function of time t in seconds. (Use -9.8 m/s2 as the acceleration due to gravity.)
h(t) =



How long will it take the rock to hit the canyon floor? (Give your answer correct to 1 decimal place.)
________s



ok so for the first part I got
h(t)= -9.8t+1100

do i take the integral of that? or what steps do i take?
Thanks
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October 22nd, 2009, 10:55 PM   #2
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Re: Differential Equation: Motion Problem

Start with acceleration:
a(t) = -9.8
Integrate this to get velocity:
v(t) = -9.8t + v0
where v0 is your initial velocity. Here the word "dropped" implies v0 = 0, so v(t) = -9.8t.
Integrate this to get position:
h(t) = -4.9t^2 + h0
where h0 is your initial height. Here, you are told it is 1100 meters, so
h(t) = -4.9t^2 + 1100

To find the time it takes the hit the canyon floor, let h(t) = 0 and find t.
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October 24th, 2009, 11:38 AM   #3
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Re: Differential Equation: Motion Problem

thank you so much for explaining...this really helped
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