Why Does Height Formula Use -16 Instead of -32?
Date: 01/31/2008 at 17:52:42 From: Andy Subject: why does the falling object model use 16 (not 32) h = -16t^2 + s is given as the position of a falling object model, but 32ft per sec^2 is the acceleration of gravity. If something is falling with the acceleration of gravity (32ft per sec^2) why to find the position do you use 16 ft per sec^2? I haven't had physics (I'm an alg II student), and I can't connect this to real life. Can you explain it?
Date: 01/31/2008 at 19:57:23 From: Doctor Rick Subject: Re: why does the falling object model use 16 (not 32) Hi, Andy. The coefficient of the t^2 term is HALF the acceleration. That's the reason for the 16. When you get to calculus, you will learn how this works (the acceleration is the second derivative of at^2, which is 2a). Before you get there, you can get the idea by constructing a table. Suppose an object starts at rest and begins falling at t=0. The instantaneous speed after one second is 32 feet per second; the *average* speed over the first second (from t=0 to t=1) is (0+32)/2 = 16 feet/sec. If the average speed over each 1-second interval increases by 32 feet per second from one second to the next, then we can tabulate the distance moved in each second, and add them up to get the position at the end of each 1-second interval: Time interval Avg. speed Dist moved Cumulative distance ------------- ---------- ---------- ------------------- 0 to 1 second 16 ft/sec 16 feet 16 feet 1 to 2 seconds 48 ft/sec 48 feet 64 feet 2 to 3 seconds 80 ft/sec 80 feet 144 feet 3 to 4 seconds 112 ft/sec 112 feet 256 feet What formula gives the position as a function of the elapsed time? It's 16 times the square of the number of seconds: after 1 second: 16 * 1^2 = 16 feet after 2 seconds: 16 * 2^2 = 64 feet after 3 seconds: 16 * 3^2 = 144 feet after 4 seconds: 16 * 4^2 = 256 feet - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 02/01/2008 at 14:32:33 From: Andy Subject: Thank you (why does the falling object model use 16 (not 32)) Thank you for the explanation, it was exactly what I needed. I still have some thinking to do about the 16 for the first second but 32 for every subsequent second, but it is starting to make more sense. Can't wait to get to calculus!
Date: 02/01/2008 at 15:26:04 From: Doctor Rick Subject: Re: Thank you (why does the falling object model use 16 (not 32)) Hi, Andy. I'll say a bit more about the 16 for the first second. You know that constant acceleration means the velocity is increasing at a constant rate. Suppose we broke it down into 1/10 second intervals. If the velocity is 0 at t=0, and 32 at t=1, then it will be as follows at the 1/10 second intervals: t v --------- 0.0 0.0 0.1 3.2 0.2 6.4 0.3 9.6 0.4 12.8 0.5 16.0 0.6 19.2 0.7 22.4 0.8 25.6 0.9 28.8 1.0 32.0 You could average all these values the long way--or you could notice that the first plus the last equals 32.0, and the second plus the next-to-last also equals 32.0, and so on. If we average the two values in any of those pairs, we'll get 32.0/2 = 16.0. Then we can average these averages; they are all the same, so the average must be 16.0. In general, for uniformly accelerated motion, the average velocity over an interval is equal to the average of the initial and final velocities. And in general (not just for uniformly accelerated motion), the distance traveled is equal to the average velocity multiplied by the time interval. That's how I built my table. Calculus will extend this thinking by breaking down the interval further and further, approaching infinitely many parts. Such thinking will then become exact, and will be put on solid ground. But for constant acceleration, it turns out that the correct result is exactly the same as what I have shown you by some slightly fuzzy arguments. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 02/01/2008 at 15:51:30 From: Andy Subject: Thank you (why does the falling object model use 16 (not 32)) Fuzzily getting clearer (a lot) - thanks for getting down to my level.
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