What does Δy=[(sin2θ)(Vi)]^2/2a represent in projectile motion?

[staka]

I would like to simply know what Δy=[(sin2θ)(Vi)]^2/2a represents..
The formula only requires the angle and the initial velocity (along with the usual a=9.8m/s^2), so what height does the equation represent?

 

[cosmo123]

The distance a projectile travels (over flat ground) is [sin(2θ)*(v^2)]/a, very similar to what you have there. Is there a chace you misread it?

 

[staka]

well that's the distance..
I think I got it now though, it's possibly the maximum height over a flat ground.

 

[cosmo123]

Oh yeah you're correct, i just googled it a bit and it came up. It is the maximum height you reach

 

[Doc Al]

I would like to simply know what Δy=[(sin2θ)(Vi)]^2/2a represents..
The formula only requires the angle and the initial velocity (along with the usual a=9.8m/s^2), so what height does the equation represent?

That's almost the formula for the maximum height of a trajectory. To correct it, replace sin2θ with sinθ.

 

[dave_baksh]

So one of the famous equations for the motion of a projectile is as follows:

v²=u²+2as

v = final velocity
u = initial velocity
a = g = -9.8m/s²
s = distance travelled

Now, for a projectile being fired at an angle, the vertical component of velocity is usin%, where % is the angle between the ground and the direction of projection.

Rearrange your equation, with v=0 to get the maximum height attained by a projectile (note that in your equation, your v is my u).

You get:

s = u²/-2a = (usin%)²/-2a

As a = -9.8, you can ignore the minus sign and you basically have your equation (except for the sin2% bit).

Yes I don't know how to make greek alphabet symbols, so a % for an angle will do :D

Hope that helps.