Projectile motion formula

  • Thread starter staka
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23
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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?
 
24
0
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?
 
23
0
well that's the distance..
I think I got it now though, it's possibly the maximum height over a flat ground.
 
24
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Oh yeah you're correct, i just googled it a bit and it came up. It is the maximum height you reach
 

Doc Al

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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.
 
21
0
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.
 

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