Rate of acceleration/gravityDate: 04/09/97 at 16:09:34 From: Catherine Mangan Subject: Rate of acceleration/gravity My question is: If a ball is dropped from a building and hits the ground after 3 seconds, how tall is the building? I think that the rate of acceleration is 9.8 meters/second or something like that. My science teacher gave us the information that at one second, the ball is traveling 9.8 meters per scond, at two seconds the ball is traveling at a rate of 19.6 meters/second, and when the ball hits the ground at three seconds, the terminal velocity was 29.4 meters per second. I would appreciate your help on this question, for I really have no idea how to solve it. Thank you. Date: 04/10/97 at 15:45:43 From: Doctor Anthony Subject: Re: Rate of acceleration/gravity Hello Catherine, Neglecting air resistance the ball will experience a constant acceleration of 9.8 m/sec. There are four very useful CONSTANT acceleration formulae which can be used for this type of problem. Any textbook on elementary mechanics will show you the derivation of these formulae, so look them up. If u = initial velocity v = final velocity s = distance a = acceleration t = time The formulae are s = (1/2)(u+v)t ..........(1) v = u + at ..............(2) s = ut + (1/2)at^2 ........(3) v^2 = u^2 + 2as ..........(4) In the problem you gave we have u = 0 a = 9.8 t = 3 s = ? Knowing u, a, and t, formula (3) will give us the value of s (= height of building) s = 0 x 3 + (1/2) x 9.8 x 9 = 44.1 metres. If you want the terminal velocity use formula (2) v = u + at = 0 + 9.8 x 3 = 29.4 m/sec -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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