Ballistic projectile

[pavelcherepan]

Howdy all! I've ran into a slight issue and I hope you guys can clarify where I went wrong.

 

Basically there I was today sitting at some boring meeting barely listening and out of boredom was trying to derive formulas for ballistic projectile trajectory. Unfortunately, when meeting finally finished and I went into wiki and compared what I have to what the actual formulas are mine were incorrect and I've ran some numbers and I definitely get wrong results. It's all very embarrassing, but I've checked my scribbles and all seems reasonable.

 

Anyways, I went like this:

 

At point 1 on the surface level we have a projectile launched with a velocity [latex]v_1[/latex] at an angle [latex]\alpha[/latex].

 

At point 2 the projectile has reached maximum height and about to start falling back down.

 

By energy conservation we have [latex]E_{k1} + E_{p1} = E_{k2} + E_{p2}[/latex]. Obviously, at point 1 projectile being at surface level the potential energy is 0, then:

 

[latex]\frac{mv_1^2}{2} = \frac{mv_2^2}{2} + mgh[/latex]

 

Dividing both sides by m and multiplying by 2 we get [latex]v_1^2 = v_2^2 + 2gh[/latex], but since the vertical component of the original [latex]v_1[/latex] vector is now 0, [latex]v_2[/latex] is essentially the horizontal component of the starting velocity or [latex]v_1*cos(\alpha)[/latex]. Then,

 

[latex]v_1^2 = v_1^2*cos^2(\alpha)+2gh[/latex]

 

Simplifying that I got:

 

[latex]h = \frac{v_1^2(1-cos^2(\alpha))}{2g}[/latex], or [latex]h = \frac{v_1^2sin^2(\alpha)}{2g}[/latex]

 

After that I went to derive the range of projectile. Since two parts of trajectory, namely, surface to highest point and highest point back to surface are essentially mirror images of one another I got:

 

[latex]R = v_{1h}*2t [/latex], where [latex]v_{1h}[/latex] is the horizontal component of the starting velocity and [latex]t[/latex] is the time required for projectile to get to the highest point of trajectory.

 

At point 2 vertical velocity is 0, then:

 

[latex]v_{1v} - gt^2/2 = 0[/latex]

 

simplifying for t,

 

[latex]t = \sqrt{\frac{2v_{1v}}{g}}[/latex], but since [latex]v_{1v} = v_1*sin(\alpha)[/latex]

 

[latex]t = \sqrt{\frac{2v_{1}*sin(\alpha)}{g}}[/latex]

 

And then range becomes:

 

[latex]R=v_{1h}*2*\sqrt{\frac{2v_{1}*sin(\alpha)}{g}} = 2v_{1}cos(\alpha)\sqrt{\frac{2v_{1}*sin(\alpha)}{g}}[/latex]

 

And this is all wrong. I've definitely done some very stupid mistake but for the life of me can't find where exactly. Any ideas?

Edited October 14, 2015 by pavelcherepan

[swansont]

 

[latex]R = v_{1h}*2t [/latex], where [latex]v_{1h}[/latex] is the horizontal component of the starting velocity and [latex]t[/latex] is the time required for projectile to get to the highest point of trajectory.

 

At point 2 vertical velocity is 0, then:

 

[latex]v_{1v} - gt^2/2 = 0[/latex]

 

 

Why is this set equal to zero? And do the units work?

[studiot]

 

At point 2 vertical velocity is 0, then:

 

 

 

This is incorrect.

 

t is not squared in the formula final velocity = initial velocity + acceleration times time (V = U +ft)

 

Be careful with your signs as well is g positive or negative?

 

the correct expression for time to vertex is

 

t =( v₁sin(a))/2

Edited October 14, 2015 by studiot

[pavelcherepan]

Oy vey! I knew it was some very stupid mistake. [latex]\frac{gt^2}{2}[/latex] is the distance traveled by free-falling body, not sure why I used it here.

 

Thanks for the help guys!

[imatfaal]

Oy vey! I knew it was some very stupid mistake. [latex]\frac{gt^2}{2}[/latex] is the distance traveled by free-falling body, not sure why I used it here.

 

Thanks for the help guys!

 

it looked like a bad mixture of v = u+at and s=ut +1/2 at^2. As per both above posters; whenever you reach an impasse like this check your dimension a/o units whichever you feel most comfortable with

 

your equation was

ms^-1 = ms^-2 s² or [length] [time]^-1 = [length] [time]^-2 [time]²

which are clearly wrong