Recognitions:
Homework Help

## Objective questions regarding "projectile motion - 2 dimension".

 Quote by sankalpmittal Ok , I agree. But I can also use the equation of range of projectile on an inclined plane. R = u2sin2(β-θ)/gcosθ - gt12/2 And t1 = 2usin(β-θ)/gcos(θ) Like that , I can also express the range in terms of t2. Which method will be easy ?
Neither one, I guess. I solved the problem with my method, if you do not like it try to solve the other way.

ehild

Recognitions:
Gold Member
 Quote by 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
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 ?
 Recognitions: Homework Help I do not know what you are doing. What are y and z?????? ehild

Recognitions:
Gold Member
 Quote by ehild I do not know what you are doing. What are y and z?????? ehild
One projectile makes angle yo with line AB. So it makes angle (θ + y) with x axis.
Other projectile makes angle zo with line AB. So it makes angle (θ + z ) with x axis.

Note : Line AB itself makes angle θ (in degrees) with x axis.
 Recognitions: Homework Help 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, $$\tan(\phi_1)=\frac{v^2 +\sqrt{v^4-g^2 X^2-2v^2 gY}}{gX}$$$$\tan(\phi_2)=\frac{v^2 -\sqrt{v^4-g^2 X^2-2v^2 gY}}{gX}$$ The times of flight are $$t_1=\frac{X}{v\cos(\phi_1)}$$ and $$t_2=\frac{X}{v\cos(\phi_2)}$$. Proceed. ehild

Recognitions:
Gold Member
 Quote by 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, $$\tan(\phi_1)=\frac{v^2 +\sqrt{v^4-g^2 X^2-2v^2 gY}}{gX}$$$$\tan(\phi_2)=\frac{v^2 -\sqrt{v^4-g^2 X^2-2v^2 gY}}{gX}$$ The times of flight are $$t_1=\frac{X}{v\cos(\phi_1)}$$ and $$t_2=\frac{X}{v\cos(\phi_2)}$$. Proceed. ehild
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 ! :(
 Recognitions: Homework Help Check your parentheses. So you need to expand the terms X2/(v2cos2φ). Using that cos2φ=1/(1+tan2φ, $$\frac{x^2}{v^2\cos^2(\phi)}=\frac{x^2}{v^2} \left(1+\tan^2(\phi)\right)$$ Substituting the expression $$\tan(\phi)=\frac{v^2\pm C}{gX}$$ You get $$1+\tan^2(\phi)=\frac{v^4\pm 2 v^2C+(v^4-g^2X^2-2v^2gY)+g^2X^2}{g^2X^2}=\frac{2v^4\pm 2 v^2C-2v^2gY}{g^2X^2}$$ Now substitute $$1+\tan^2(\phi_1)=\frac{2v^2(v^2+ C-gY)}{g^2X^2}$$ for t12and $$1+\tan^2(\phi_2)=\frac{2v^2(v^2- C-gY)}{g^2X^2}$$ for t22 into the expression for E. ehild
 Recognitions: Gold Member 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 ?
 Recognitions: Homework Help 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

Recognitions:
Gold Member
 Quote by 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
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.

Recognitions:
Homework Help
 Quote by sankalpmittal Ah so , t12 = 2(v2+C-gY)/g2 t22 = 2(v2-C-gY)/g2 2t1t2sinθ = 4sinθ√(v4 + g2Y2 + 2v2gY - C2)/g2
There is a sign error (shown in red).
$$2 t_1 t_2 \sin(\theta)=\frac {4 \sin(\theta)}{g^2}\sqrt{(v^2-gY+C)(v^2-gY-C)}=\frac {4 \sin(\theta)}{g^2}\sqrt{(v^4-2gYv^2+g^2Y^2-C^2)}$$

Recognitions:
Gold Member
 Quote by ehild There is a sign error (shown in red). $$2 t_1 t_2 \sin(\theta)=\frac {4 \sin(\theta)}{g^2}\sqrt{(v^2-gY+C)(v^2-gY-C)}=\frac {4 \sin(\theta)}{g^2}\sqrt{(v^4-2gYv^2+g^2Y^2-C^2)}$$
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 !)