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 March 5th, 2017, 08:51 PM #1 Senior Member   Joined: Jan 2017 From: US Posts: 104 Thanks: 5 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!
 March 5th, 2017, 08:52 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,860 Thanks: 968 assume $v_0=0$ Thanks from Indigo28
 March 11th, 2017, 01:36 PM #3 Newbie   Joined: Feb 2017 From: Omaha, Nebraska Posts: 8 Thanks: 0 Math Focus: Algebra and Trigonometry 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.
 March 11th, 2017, 02:16 PM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 $h(t) = 64-16t^2$
 March 12th, 2017, 03:00 AM #5 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 If the initial velocity has zero vertical component it plays no role.
 March 12th, 2017, 09:52 AM #6 Newbie   Joined: Feb 2017 From: Omaha, Nebraska Posts: 8 Thanks: 0 Math Focus: Algebra and Trigonometry 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
 March 12th, 2017, 10:01 AM #7 Global Moderator   Joined: Dec 2006 Posts: 18,847 Thanks: 1568 No. With h(t) = 64 - 16t², one gets h(t) = 0 when t = 2.

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