Rock Falling From BuildingDate: 9/12/96 at 18:16:21 From: Carl C. Smith Subject: Rock Falling From Building of Height... Dear Dr.Math, A rock is dropped from a tall building and hits the ground at 224 ft/sec (approximately 150 miles/hr). How long did it take for the object to fall? I tried doing the equation, which is T=D/V. I can't do that without knowing how tall the building is, or without knowing how long it takes for the rock to reach terminal velocity. If you could write back and send some tips I would really appreciate it. Sincerely, Amber Date: 9/12/96 at 18:56:8 From: Doctor Tom Subject: Re: Rock Falling From Building of Height... Hi Amber, The problem is that you are using the wrong formula. The one you're using assumes that the rock is moving at the same speed all the time, but for your problem, it starts not moving (the instant before you let go), and it winds up moving very fast indeed. You need two formulas (which you can probably find in your book somewhere). One says that the speed at any time is: V = At, where V is the speed, A is the acceleration of gravity (maybe it's called "g"), and t is the time. You know v when it hits (224 ft/sec) and you know A (32 ft/sec^2). So you can work out t = V/A = 224/32 = 7 seconds. The other formula you need tells how far a rock will fall, starting from standing still, under constant acceleration. That formula is: D = 1/2(At^2) D will be the height of the building, A is 32, and t (you just worked it out) is 7. So D = 32*49/2 = 784 feet. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 9/15/96 at 20:52:20 From: Carl C. Smith Subject: Re: Rock Falling From Building of Height... Doctor Tom, Thanx so much for your help, I really appreciate it. Sincerely, Amber |
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