Objective questions regarding "projectile motion - 2 dimension".

ehild

Neither one, I guess. I solved the problem with my method, if you do not like it try to solve the other way.

ehild

sankalpmittal

E = [X/cos(θ+z)]² + [X/cos(θ+y)]² + 2X²sinθ/cos(θ+z)cos(θ+y)

On solving , I get :

E = (X²/(cos²θcos²z + sin²θsin²z - 2cosθcoszsinθsinz)) + (X²/(cos²θcos²y + sin²θsin²y - 2cosθcosysinθsiny)) + 2X²sinθ/(cos²θcoszcosy - sinθcosθcoszcosy - sinθsinzcosθcosy + sin²θ sinzsiny)

Am I going correct ?

ehild

I do not know what you are doing. What are y and z??????

ehild

sankalpmittal

One projectile makes angle y^o with line AB. So it makes angle (θ + y) with x axis.
Other projectile makes angle z^o with line AB. So it makes angle (θ + z ) with x axis.

Note : Line AB itself makes angle θ (in degrees) with x axis.

ehild

Do you know y and z?

I do not think expanding the expression of E had any sense. Introducing the angles with respect to AB does not make the solution easier.

You know the two angles with respect to the horizontal in terms of (X,Y), coordinates of the point to be hit by the projectile. Y=Xtan(θ). If v is the initial speed,

[tex]\tan(\phi_1)=\frac{v^2 +\sqrt{v^4-g^2 X^2-2v^2 gY}}{gX}[/tex][tex]\tan(\phi_2)=\frac{v^2 -\sqrt{v^4-g^2 X^2-2v^2 gY}}{gX}[/tex]

The times of flight are

[tex]t_1=\frac{X}{v\cos(\phi_1)}[/tex]
and
[tex]t_2=\frac{X}{v\cos(\phi_2)}[/tex].

Proceed.

ehild

sankalpmittal

Ok , so on expanding and simplifying this expression ,

E = (X²)(2g²X² + 2v⁴ + 2C)/(v²)(g²X²) + 2sinθ √(((Z+2v²√C)/g²X²))(Z-2v²√C)/(g²X²))

where ,

Z = g²x² + v⁴ + C
C = v⁴−g²X²−2v²gY

Don't know how to latex ! :(

ehild

Check your parentheses.

So you need to expand the terms X²/(v²cos²φ). Using that cos²φ=1/(1+tan²φ,

[tex]\frac{x^2}{v^2\cos^2(\phi)}=\frac{x^2}{v^2} \left(1+\tan^2(\phi)\right)[/tex]

Substituting the expression
[tex]\tan(\phi)=\frac{v^2\pm C}{gX}[/tex]
You get
[tex]1+\tan^2(\phi)=\frac{v^4\pm 2 v^2C+(v^4-g^2X^2-2v^2gY)+g^2X^2}{g^2X^2}=\frac{2v^4\pm 2 v^2C-2v^2gY}{g^2X^2}[/tex]

Now substitute
[tex]1+\tan^2(\phi_1)=\frac{2v^2(v^2+ C-gY)}{g^2X^2}[/tex]
for t₁²and
[tex]1+\tan^2(\phi_2)=\frac{2v^2(v^2- C-gY)}{g^2X^2}[/tex]
for t₂²
into the expression for E.

ehild

sankalpmittal

Ah so ,

t₁² = 2(v²+C-gY)/g²

t₂² = 2(v²-C-gY)/g²
2t₁t₂sinθ = 4sinθ√(v⁴ + g²Y² + 2v²gY - C²)/g²

So Expression E =

[2(v²+C-gY) + 2(v²-C-gY) + 4sinθ√(v⁴ + g²Y² + 2v²gY - C²)]/g²

Ok , so how to prove this expression really is independent of θ ?

ehild

You see that +C cancels with -C in the first part. As for the expression under the square root, substitute C² = v⁴−g²X²−2v²gY and simplify.

ehild

sankalpmittal

Allrighty.

So simplifying E = [2(v²+C-gY) + 2(v²-C-gY) + 4sinθ√(v⁴ + g²Y² + 2v²gY - C²)]/g² further , I get ,

E =[4v² - 4 gY + 4sinθ√(g²Y² + g²X² + 4v²gY)]/g²

tanθ = Y/X

So , expression E reduces to ,

E =[4v² - 4 gXtanθ + 4sinθ√(g²X²tan²θ + g²X² + 4v²gXtanθ)]/g²

Which again is simplified to ,

E =[4v² - 4 gXtanθ + 4sinθ√(g²X²sec²θ + 4v²gXtanθ)]/g²

Shall I continue ahead ? Please tell if I'm on correct path. Thanks.

ehild

There is a sign error (shown in red).
[tex]2 t_1 t_2 \sin(\theta)=\frac {4 \sin(\theta)}{g^2}\sqrt{(v^2-gY+C)(v^2-gY-C)}=\frac {4 \sin(\theta)}{g^2}\sqrt{(v^4-2gYv^2+g^2Y^2-C^2)}[/tex]

sankalpmittal

Ah ! Well spotted !

So , all I've gotta do is to simplify the expression E = [2(v²+C-gY) + 2(v²-C-gY) + 4sinθ√(v⁴ + g²Y² - 2v²gY - C²)]/g²

Thus I reduce this expression by putting C² = v⁴−g²X²−2v²gY to

E =[4v² - 4 gY + 4sinθ√(g²Y² + g²X² )]/g²

Again substituting Y/X for tanθ , expression E reduces to ,

[4v² - 4 gXtanθ + 4sinθ√(g²X²tan²θ + g²X² )]/g²

Which again is reduced to ,

[4v² - 4 gXtanθ + 4sinθ. gXsecθ]/g²

Which is simplified again to ,

[4v² - 4 gXtanθ + 4gXtanθ]/g²

And lastly we get ,

4v²/g² which is of course independent of θ !

Thanks ehild ! Thanks for efforts !

(I'll post some problems of friction soon !)