anonymous
  • anonymous
What is the free fall equation?
Physics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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theEric
  • theEric
There are a few basic free fall equations that ignore air resistance and changing gravity. They are basic motion equations, assuming a constant acceleration (constant force), and that acceleration is gravitational acceleration (symbolized as \(g\)). One very helpful one is \(x=x_0+v_0\ \Delta t+\frac{1}{2}g\ \Delta t^2\) where \(\Delta t=t-t_0\) Subscripts of \(0\) are for the initial value, while the absense of a \(0\) means it is the second, or current value. \(x\equiv\quad\)displacement \(v\equiv\quad\)velocity \(t\equiv\quad\)time
theEric
  • theEric
If you know that acceleration is \(a=\dfrac{v-v_0}{t-t_0}\), then you'll know that with \(a=g\) then \(g=\dfrac{v-v_0}{t-t_0}\). So you can see things like \(v=g(t-t_0)+v_0\). \(v^2=v_0^2+2\ a\ \Delta x\) is another useful equation. Any questions? Is that what you asked for?
Isaiah.Feynman
  • Isaiah.Feynman
Well to summarize what theEric said, there are four equations for linear motion, at first we use these equations for an object whose motion is only horizontal. In free fall we modify these equations a little and then apply to an object whose motion is only vertical.

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oOKawaiiOo
  • oOKawaiiOo
Dude, take the shortcut. Follow me...... This only applies when initial V = 0 \[\sqrt{2*gravity*height}= velocity\] or \[-5t^{2} = distance\] Gravity is always : 10 m/s^2 Height is how far it is away from the ground. Plug in the values.
theEric
  • theEric
@oOKawaiiOo , I think that gravity is a force, and is not something that you want to calculate. Did you mean gravitational acceleration? So \(\sqrt{2\ g\ d}=v\)? http://en.wikipedia.org/wiki/Equations_for_a_falling_body Also, assuming \(x_0=v_0=0\) and \(a=g\), then \(\ x=x_0+v_0\ t+\frac{1}{2}g\ t^2\\\quad=\frac{1}{2}g\ t^2\\\implies t^2=2\ g\ x\\\implies -5t^2=-10\ g\ x\\\quad = x\ \text{if }-10g=1\). Gravitational acceleration is closer to \(9.82\ [m/s^2]\) than \(10\ [m/s^2]\), but the teacher will decide what approximation of \(g\) should be used. Assuming either value for \(g\), \(-10g\neq 1\) and so \(-5t^2\neq x\). So was that a typo?
oOKawaiiOo
  • oOKawaiiOo
If air resistance is ignored, you can use the approximation of 10m/s^2 Gravity is an acceleration. A force, from a physics perspective, is Newton (N), which has the units of \[N= \frac{ Kg*m }{s^{2} }\] The wikipedia site you posted, both of my equations I used are on that site.
theEric
  • theEric
But I proved that \(-5t^2\neq x\) on Earth, didn't I? Also, \(g\) is used for gravitational acceleration. A force is in newtons, which is equivalent to \([kg\ m/s^2]\). But, for that equation, you want gravitational acceleration, the acceleration. This is the acceleration due to gravity, which is a force. Now, that is all semantics, but it doesn't not alter my proof! Is my proof correct? As I said, you can use different approximations for gravitational acceleration. The \(10\ [m/s^2]\) is one common approximation. I could not find your second equation at the link I posted. However, there is the first one that is \(d=\frac{1}{2}g\ t^2\), which implies that \(t^2=\dfrac{2\ d}{g}\) and so \(-5t^2=\dfrac{-10g}{d}\). Now, if it is true that \(-5t^2=\dfrac{-10g}{d}\), and that \(-5t^2=d\), then \(\dfrac{-10g}{d}=d\). This implies that \(-10g=d^2\). Since \(-10g\) is nearly constant of a falling object that we are describing, then \(d^2\) must also be constant, even with changing time and a non-zero acceleration. That is not true - the square of the distance i very obviously changing. Assuming Wikipedia's equation to be correct, and the reasoning that the square of the distance will not be equal to the opposite of the acceleration increased by a factor of 10, then your equation is incorrect. Which assumption of mine would be wrong? I just don't want a flawed equation to be on OpenStudy, so please show me the error in my earlier proof. This last one is not an actual proof, because it is true only assuming that \(-10g=d^2\). This is because gravitational acceleration changes with distance, though, and it takes a considerable distance to make a slight change in gravitational acceleration.
oOKawaiiOo
  • oOKawaiiOo
"gravitational acceleration changes with distance" No its doesnt. Gravity is a constant value, constant acceleration no matter at what height you drop it. The final speed is affected by the height dropped. PE = KE mgh = 0.5mv^2 if you solve for V the equation is \[\sqrt{2*gravity*height}=velocity\] This equation is the 4th equation down on wikipedia.
Isaiah.Feynman
  • Isaiah.Feynman
@oOKawaiiOo Gravitational acceleration DOES decrease as altitude increases.
theEric
  • theEric
Yes. See, \(g\approx G\dfrac{M_\text{Earth}}{r^2}\) where \(M\) is mass, \(r\) is distance between the objects' centers of gravity, and \(G\) is the gravitational constant. And I am not debating the \(\sqrt{2gh}=v\) equation, I agree! You made a good point! \(PE_1+KE_1=PE_2+KE_2\) assuming that energy is conserved in the closed system of the object and Earth. I used subscript \(1\) for the first point in time, and subscript \(2\) for the second. If you fall from rest, \(KE_2=0\). You will probably travel the entire distance \(h\) because it is common to \(\sf\color{green}{choose}\) \(h\) to be the distance you travel. Then, at the second point, \(PE_2=0\). The result is \(PE_1=KE_2\\\qquad =m_1g_1h=\frac{1}{2}m_2v_2^2.\) Assuming \(m_1=m_2\), then \(\sqrt{2gh}=v\).

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