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 June 7th, 2011, 12:06 PM #1 Member   Joined: Mar 2011 Posts: 49 Thanks: 1 Dimensional analysis "From the dimensional analysis find the time t it takes for a ball to fall from a height h. " Anyone able to explain me exactly what this is asking for?
 June 7th, 2011, 12:39 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Dimensional analysis Here is an article on dimensional analysis. I suppose we could start with: $t=\frac{h}{\bar{v}}$ We have on the left a unit of time and on the right length divided by length per time which is time so it is dimensionally consistent. For free-fall (ignoring drag and other non-conservative forces), we have: $\bar{v}=\frac{v_i+v_f}{2}$ Since the ball is said to fall, we assume $v_i=0$ and with constant acceleration we have $v_f=gt$ where we take down to be in the positive direction, thus: $\bar{v}=\frac{1}{2}gt$ Notice on the left we have units of length per time and on the right we have length per time squared times time = length per time. Now substituting for $\bar{v}$ we get: $t=\frac{h}{\frac{1}{2}gt}=\frac{2h}{gt}$ $t^2=\frac{2h}{g}$ $t=\sqrt{\frac{2h}{g}}$ Notice we have units of time on both sides. Notice the expression agrees with intuition as well, the greater the value of h the greater the value of t while the greater the value of g, the smaller the value of t.
 June 7th, 2011, 01:47 PM #3 Member   Joined: Mar 2011 Posts: 49 Thanks: 1 Re: Dimensional analysis The answer you wrote makes perfect sense to me, but the solution that was given to this question does not.. xD They give Solution: We know that the time should depend on the height, h, and free fall acceleration, g. So we can write: $t= ah^xg^y$ Where a is a dimensionless constant (we cannot find this constant) and x, y – are constant, which can be found from dimensional analysis. The units of t is seconds. The units of h is meter, the unit of g is . Then we have:  From this equation we have: $1=-2y 0 = x+y$ Then $y=-1/2 x=-y=1/2$ Then $t=a(h/g)^1/2$
 June 7th, 2011, 02:05 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Dimensional analysis Okay, I see what they've done, and their answer agrees with what I gave where $a=\sqrt{2}$ They correctly posit that the time will be a function of h and g, where both have rational exponents: $t=ah^xg^y$ Now, with t having units of seconds, h has units of meters and g has units of meters/seconds˛, dimensional analysis yields: $\text{s}=a\text{m}^x$$\text{\frac{m}{s^2}}$$^y$ $\text{s}^1\text{m}^0=a\cdot\text{m}^{x+y}\text{s}^ {-2y}$ This means: $x+y=0$ $-2y=1\:\therefore\:y=-\frac{1}{2}$ $x-\frac{1}{2}=0\:\therefore\=\frac{1}{2}" /> thus: $t=ah^{\frac{1}{2}}g^{-\frac{1}{2}}=a\sqrt{\frac{h}{g}}$
 June 7th, 2011, 05:11 PM #5 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0
 June 8th, 2011, 06:53 AM #6 Member   Joined: May 2011 Posts: 90 Thanks: 1 Re: Dimensional analysis Hello MofD Time is measured in seconds We cannot have "apples" = "oranges" So whatever you say is time must be seconds not height, speed or litres. OK So if you want time to eaqual some combination of height and g you are asking to get seconds from height and acceleration acceleration is distance per second each second and height is distance So gh would be (distance per second) squared So root gh would be a speed And distance is speed times time If you wrote all that, you'd get the marks.
 June 9th, 2011, 08:47 AM #7 Member   Joined: Mar 2011 Posts: 49 Thanks: 1 Re: Dimensional analysis I understand with it now, after quarreling a bit with it.

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