Which Physics Formula To Use

[dmission]

Hi All,

I'm having a tough time distinguishing between these two formulas:
x=(1/2)*a*t^2
and
v=sqrt(2gh)

It seems like both can be used to find height. In the first one, x would be solved for, and in the second one, if you knew x and time, then you know V, and could solve for H? However, in the problem I used where you're trying to find a projectile's initial height off of the ground, only the first equation works, not the second one. Is there any way to distinguish when these two can be used?

Thanks

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[exquisitemelody]

Do you have an example of when you're trying to use the second equation and it doesn't work? Here's my theory...

I'm assuming you're talking about a problem where an object is either dropped of the cliff or thrown off the cliff and you're trying to find the height of the cliff?

If you're simply dropping the object off the cliff, you can use the second equation, because that equation is from conservation of energy, correct? Therefore, the initial potential energy of the object will equal the kinetic energy of the object. This requires knowing the final velocity.

However, if you're throwing the object off the cliff, you can't use the second equation because the object has some initial kinetic energy (versus the first case where you don't have initial kinetic energy). This initial kinetic energy isn't taken into account when you use that second equation.

 

[Arctangent]

The two equations basically can be derived from each other, since the second equation:
Vf = sqrt(2gh)
Vf^2 = 2gh
Vf * Vf/g = 2h
(assuming Vo = 0)
Vf * t = 2h
1/2 Vf * t = h
(assuming g is a constant)
Vavg * t = h

First equation:
x = (1/2)at^2
x = (1/2) * (at) * t
(assuming Vo is 0)
x = (1/2) * Vf * t
(assuming a is a constant)
x = Vavg * t

So as long as you use the formulas correctly, there shouldn't be an issue with either. Could you post an example question or something if you have it?

 

[MShopes]

Hi All,

I'm having a tough time distinguishing between these two formulas:
x=(1/2)*a*t^2
and
v=sqrt(2gh)

It seems like both can be used to find height. In the first one, x would be solved for, and in the second one, if you knew x and time, then you know V, and could solve for H? However, in the problem I used where you're trying to find a projectile's initial height off of the ground, only the first equation works, not the second one. Is there any way to distinguish when these two can be used?

Thanks

Both equations are almost the same thing. The only difference is one has the final velocity while the other has the time variable. So use whichever you need depending on the question.

Both equations only apply to objects dropped because the initial velocity is O.
For the first equation, the formula is X=Vot+1/2 at^2, with Vo(initial velocity)=0, we cancel Vot and get X=1/2 at^2.

For the second equation, the formula is V^2=Vo^2+2ax, with Vo=0, we can use cancel Vo^2 and thus get V^2=2ax, or V= squareroot (2ax), x is same as h I just used different symbol. And a is basically g however a in this case. The only acceleration to projectile motion objects is g.

 

[dmission]

Thanks for all the replies.

Do you have an example of when you're trying to use the second equation and it doesn't work? Here's my theory...

I'm assuming you're talking about a problem where an object is either dropped of the cliff or thrown off the cliff and you're trying to find the height of the cliff?

If you're simply dropping the object off the cliff, you can use the second equation, because that equation is from conservation of energy, correct? Therefore, the initial potential energy of the object will equal the kinetic energy of the object. This requires knowing the final velocity.

However, if you're throwing the object off the cliff, you can't use the second equation because the object has some initial kinetic energy (versus the first case where you don't have initial kinetic energy). This initial kinetic energy isn't taken into account when you use that second equation.

It seems to me like this is probably the closest response to being correct. I've tried both formulas to solve a problem, and only the x=1/2at^2 yields the correct answer.

The question, roughly, is this:
a projectile is shot from a gun, and the gun is located on the hill. Given that the projectile's final displacement was 14m, and it travelled that far in 1.4seconds (v=10m/s?), how high above the ground was the projectile?
Using x = 1/2(10)(1.4^2) yields the correct answer, but I think that 10=sqrt(2*10*h) does not. Would appreciate any clarification, thanks.

 

[Arctangent]

Thanks for all the replies.
It seems to me like this is probably the closest response to being correct. I've tried both formulas to solve a problem, and only the x=1/2at^2 yields the correct answer.

The question, roughly, is this:
a projectile is shot from a gun, and the gun is located on the hill. Given that the projectile's final displacement was 14m, and it travelled that far in 1.4seconds (v=10m/s?), how high above the ground was the projectile?
Using x = 1/2(10)(1.4^2) yields the correct answer, but I think that 10=sqrt(2*10*h) does not. Would appreciate any clarification, thanks.

The formula V = sqrt(2gh) is more specifically Vfinal = sqrt(2gh), which is important for this problem. You are given Vinitial, and also in the x/horizontal direction. Thus, if you were to use this equation, you would need to find Vfinal. As the bullet travels for 1.4s, the initial vertical component of velocity is 0m/s (the bullet is shot horizontally I presume), and the acceleration due to gravity is 10m/s^2:
Vf = Vi + at
Vf = gt
Vf = 10m/s * 1.4s
Vf = 14m/s

If you plug this in:
h = Vf^2/(2g)
h = (14m/s)^2/(20m/s^2) = 9.8m
Which is the same answer as if you used x = (1/2) at^2

The two equations basically differ in that one uses final velocity, and the other uses time. Since you are given time and not final speed, you should opt to use x = (1/2) at^2 in the interest of time.

 

[cge0]

...

 

[dmission]

The formula V = sqrt(2gh) is more specifically Vfinal = sqrt(2gh), which is important for this problem. You are given Vinitial, and also in the x/horizontal direction. Thus, if you were to use this equation, you would need to find Vfinal. As the bullet travels for 1.4s, the initial vertical component of velocity is 0m/s (the bullet is shot horizontally I presume), and the acceleration due to gravity is 10m/s^2:
Vf = Vi + at
Vf = gt
Vf = 10m/s * 1.4s
Vf = 14m/s

If you plug this in:
h = Vf^2/(2g)
h = (14m/s)^2/(20m/s^2) = 9.8m
Which is the same answer as if you used x = (1/2) at^2

The two equations basically differ in that one uses final velocity, and the other uses time. Since you are given time and not final speed, you should opt to use x = (1/2) at^2 in the interest of time.

Thanks for the reply. How do you know, though, that Vi is 0? Wouldn't that be the speed it's shot out of the gun at?

 

[exquisitemelody]

Thanks for the reply. How do you know, though, that Vi is 0? Wouldn't that be the speed it's shot out of the gun at?

Velocity is a vector with a horizontal and vertical component. The object is initially shot horizontally, which means there's no initial vertical velocity. The final velocity the other poster found was the final vertical velocity (btw, horizontal velocity is constant).

 

[dmission]

In my physics book, they say that v=sqrt(2gh) can be used only if Vi or Vf is 0, would that still apply in this case? Why is the velocity not 10m/s, if it travelled 14m in 1.4 seconds?

 

[Arctangent]

In my physics book, they say that v=sqrt(2gh) can be used only if Vi or Vf is 0, would that still apply in this case? Why is the velocity not 10m/s, if it travelled 14m in 1.4 seconds?

Horizontal motion and vertical motion need to be differentiated. Whenever you are solving for something along the lines of flight time/vertical height between points/etc., you will look only at the vertical component of motion. If the bullet is shot horizontally, then by definition, as it leaves the gun, there is a horizontal component of velocity, but there is no vertical component. The additional information (14m in 1.4s) was there to give you time of flight and to feed you extraneous (horizontal) information that is not necessary for this particular problem. The initial vertical velocity is 0m/s, so v=sqrt(2gh) can be used, you just need to be able to discern that initial vertical velocity is what is needed, and that that quantity is 0m/s.

 

[dmission]

Horizontal motion and vertical motion need to be differentiated. Whenever you are solving for something along the lines of flight time/vertical height between points/etc., you will look only at the vertical component of motion. If the bullet is shot horizontally, then by definition, as it leaves the gun, there is a horizontal component of velocity, but there is no vertical component. The additional information (14m in 1.4s) was there to give you time of flight and to feed you extraneous (horizontal) information that is not necessary for this particular problem. The initial vertical velocity is 0m/s, so v=sqrt(2gh) can be used, you just need to be able to discern that initial vertical velocity is what is needed, and that that quantity is 0m/s.

Do we want the initial vertical velocity because we're solving for height? I'm a bit confused still about a this though: wouldn't the initial velocity be the speed when it's shot out?

Secondly, velocity = displacement/time, so what velocity was I calculating when I did v = 14m/1.4sec? Not initial or final?

 

[Arctangent]

Do we want the initial vertical velocity because we're solving for height? I'm a bit confused still about a this though: wouldn't the initial velocity be the speed when it's shot out?

Secondly, velocity = displacement/time, so what velocity was I calculating when I did v = 14m/1.4sec? Not initial or final?

Because you are solving for height (which implies vertical distance) you need initial vertical velocity. Initial velocity is the speed/direction when it's shot out, but initial vertical velocity is the what you need (because you are concerned with height).

The velocity you calculated was horizontal velocity. In MCAT format, horizontal velocity does not change, so technically you found initial/final horizontal velocity, which does not matter for solving for height.

 

[Danlee07]

the answer's been answered, but i'd like to share a tip... do as many problems that are easily distinguishable as vector component problems. You want to get a good understanding of the vectors... it's the basis for the rest of your physics. relate everything back to vectors. eventually, you'll be able to freely use terms such as velocity and terminal velocity without confusion.