Basic Conversion from Translational Motion Equation

[TrueBlood]

Studying for the MCAT, and trying to figure out how t = sqrt(2h/g) and t = Vo*sqrt(2h/g) is derived from the standard translational motion equations.

h = change in h
t = change in t

thus...

h = volt + 1/2g(t)^2
h - volt = 1/2g(t)^2
(2h)/g - (2Vot)/g =t^2
sqrt(2h/g) - sqrt(2Vot)/g = t

Obviously, I don't know how to simplify it further.

Thanks.

 

[Doc Al]

Studying for the MCAT, and trying to figure out how t = sqrt(2h/g) and t = Vo*sqrt(2h/g) is derived from the standard translational motion equations.

Those equations cannot both be correct--they have different dimensions!

Only the first is correct, and only when Vo = 0.

 

[Doc Al]

h = volt + 1/2g(t)^2
h - volt = 1/2g(t)^2
(2h)/g - (2Vot)/g =t^2
sqrt(2h/g) - sqrt(2Vot)/g = t

Also realize that your last step is mathematically incorrect:

\sqrt{a^2 - b^2} \ne a - b

For example sqrt(5^2 - 3^2) = sqrt(16) = 4 ≠ 5 - 3

 

[TrueBlood]

The last step, mathematically correct, would be...

sqrt[(2h-2Vot)/g] = t

and in a free-fall equation, where Vo = 0, then t = sqrt(2h/g)

I found out t = Vo*sqrt(2h/g) when you are trying to find the range (r = volt) and using t = sqrt(2h/g) when you don't have t.
Thx.

 

[Doc Al]

The last step, mathematically correct, would be...

sqrt[(2h-2Vot)/g] = t

and in a free-fall equation, where Vo = 0, then t = sqrt(2h/g)

Good.

I found out t = Vo*sqrt(2h/g) when you are trying to find the range (r = volt) and using t = sqrt(2h/g) when you don't have t.

Careful. That equation as written makes no sense. You probably meant to use r instead of t. And t = sqrt(2h/g) would only be half the time for the full trajectory.