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March 5th, 2017, 09:51 PM   #1
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Question I don't think there is enough information here...

At a diving competition, divers jump off a 64 foot high cliff. Solve for how long it will take the diver to hit the water. Show your work and explain the steps you used to solve.

Correct me if I'm wrong, but I feel like you would need more information (i.e., the velocity of the dive) to solve this. If I'm wrong could someone please help me with this?

Thanks in advance!
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March 5th, 2017, 09:52 PM   #2
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assume $v_0=0$
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March 11th, 2017, 02:36 PM   #3
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Think this might be what your wanting to do:

h(t) = 16t^2

where h is height and t^2 is time squared,
so plug in the dive distance of 64 feet and solve for t.
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March 11th, 2017, 03:16 PM   #4
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$h(t) = 64-16t^2$
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March 12th, 2017, 03:00 AM   #5
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If the initial velocity has zero vertical component it plays no role.

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March 12th, 2017, 09:52 AM   #6
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Here's the answer I get...

With initial velocity of zero, assuming zero resistance and seconds for time:
(Diver accelerates due to constant force of gravity.)

h(t) = 16t^2

64t = 16 t^2
64/16 = t^2 / t
4 = t

Last edited by y2kevin; March 12th, 2017 at 09:58 AM. Reason: add note about diver and gravity
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March 12th, 2017, 10:01 AM   #7
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No. With h(t) = 64 - 16t², one gets h(t) = 0 when t = 2.
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