Objective questions regarding projectile motion - 2 dimension .

[sankalpmittal]

Objective questions regarding "projectile motion - 2 dimension".

Homework Statement

There are 2 mini problems :

1. If t₁ and t₂ be the times of flight from A to B and "θ" be the angle of inclination of AB to the horizontal , then t₁² + 2t₁t₂sinθ + t₂² is

(A.) independent of θ
(B.) equal to 0
(C.) dependent on θ
(D.) none of the above is correct.

2. If B has horizontal and vertical coordinates x,y referred to A , and the velocity of projection is √(2gh) , and the angle between the two paths at B is a right angle , then B lies on the ellipse ,

(A.) x²+2y² = 2hy
(B) x² + y² = 2hy
(C) 2x² + 2y² = 2hy
(D) x²+2y² = 2hx

Homework Equations

http://en.wikipedia.org/wiki/Trajec...tions_at_the_final_position_of_the_projectile

The Attempt at a Solution

1. Answer seems to be C but its not !

2. I don't understand the question seriously. Langauge seems confusing to me !

Please help...
Thanks in advance ..

 

[ehild]

Homework Statement

There are 2 mini problems :

1. If t₁ and t₂ be the times of flight from A to B and "θ" be the angle of inclination of AB to the horizontal , then t₁² + 2t₁t₂sinθ + t₂² is

(A.) independent of θ
(B.) equal to 0
(C.) dependent on θ
(D.) none of the above is correct.

2. If B has horizontal and vertical coordinates x,y referred to A , and the velocity of projection is √(2gh) , and the angle between the two paths at B is a right angle , then B lies on the ellipse ,

(A.) x²+2y² = 2hy
(B) x² + y² = 2hy
(C) 2x² + 2y² = 2hy
(D) x²+2y² = 2hx

Homework Equations

http://en.wikipedia.org/wiki/Trajec...tions_at_the_final_position_of_the_projectile

The Attempt at a Solution

1. Answer seems to be C but its not !

2. I don't understand the question seriously. Langauge seems confusing to me !

Please help...
Thanks in advance ..

I do not understand the first problem at all. What are t1 and t2? Are they times of flight of two projectiles, fired at different angles or different initial speeds or both?

ehild

 

[CWatters]

Deleted.

 

[sankalpmittal]

I do not understand the first problem at all. What are t1 and t2? Are they times of flight of two projectiles, fired at different angles or different initial speeds or both?

ehild

Hey ehild ! Thanks for quick reply. :)

At first I did not understand the problem as language seemed confusing (is). However what the first question says has been illustrated in the below "attatchment :"

 

[ehild]

I still do not understand. The flight time from point A to B is a single time, not two. So again, what are those times? ehild

 

[sankalpmittal]

I still do not understand. The flight time from point A to B is a single time, not two. So again, what are those times?


ehild

Question says that :

Suppose a body is projected along a parabolic path from the point defined by (0,0). Imagine the x-axis as time axis and y-axis as position axis. From that (0,0) point , the body reach the point (t₁,A). Obviously to reach (t₁,A) , the body takes time t₁. Now to reach point (t₂,B) from here , the body takes time t₂ if the time is defined from (0,0). This means there is a time lag of t₂-t₁ between points (t₁,A) and (t₂,B).

Of course , coordinates (t₁,A) and (t₂,B) lie on a parabola. See the attatchment in above post again.

 

[sankalpmittal]

Yes, it has sense now. I guessed that it was two trajectories involved, but you said something totally different...

The second problem also means that two trajectories both start from the same point A (the origin) and cross each other at point B. The two projectiles were launched at different angles: The angles φ1 and φ2 can be obtained from the coordinates x,y of B with respect to A.
The curves are perpendicular to each other at B: That means the derivatives dy/dx are negative reciprocal to each other.

The equation of the trajectory of a projectile is
y=tan(φ)x-gx2/(2v02cos2(φ)).
You can express cos(φ) with tan(φ) and solve for tan(φ).
The equation of the derivative is dy/dx=tan(φ)-gx/(v02cos2(φ)).
Substitute the angles and then the product of the derivatives has to be -1. From that, you get an equation for x and y, and it is the equation of an ellipse.

But this is a very hard problem. Are you sure that you need to solve such problems?

I try to solve the first equation tomorrow.

ehild

Ok , but really , have I to solve that much !

Hi, sankalpmittal,

You find the formula for the Angle theta required to hit coordinate (x,y) on the Wiki-page the URL you attached to your OP.
Use it for both problems.

Ok , let's do the first problem.

Since range of two projectile is same , then

Angle of first projectile be x , then other will have 90-x.

If we join A and B , we make it a symmetric projectile.

Now how will I proceed ?

 

[sankalpmittal]

Also t₁ = 2usinx/g
t₂ = 2usin(90-x)/g = 2ucosx/g

Also see the image in the below link :

http://postimage.org/image/ptms5vrwh/

Please give some hints to proceed further. I cannot find any way out.

 

[ehild]

Ok , but really , have I to solve that much !Ok , let's do the first problem.

Since range of two projectile is same , then

Angle of first projectile be x , then other will have 90-x.

If we join A and B , we make it a symmetric projectile.

Now how will I proceed ?

That projectile is not symmetric. You want to hit a point B (X,Y) from the origin. The question is what has to be the launch angle. There are two angles,Φ₁ and Φ₂, but Φ₁ =90-Φ₂ is true only for Y=0.

The angle depends on the position of point B and on the initial speed v.
Wikipedia provides the formula

$$\tan(\phi)=\frac{v^2 \pm \sqrt{v^4-g^2 X^2-2v^2 gY}}{gx}$$http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

The time the projectile flies from the origin to B is t=X/(vcos(Φ).
You need to investigate the expression t₁²+t₂²+2t₁t₂sin(θ) where tan(θ)=Y/X.

You will find the formula 1/cos²(Φ )=1+tan²(Φ ) useful for the derivation.

ehild

 

[ehild]

Also t₁ = 2usinx/g
t₂ = 2usin(90-x)/g = 2ucosx/g

Also see the image in the below link :

http://postimage.org/image/ptms5vrwh/

Please give some hints to proceed further. I cannot find any way out.

Your equations for t are valid if the projectile final height and initial height are the same. It is not the case in this problem.

ehild

 

[ehild]

I show the solution steps a bit further:
The problem: you want to hit a point B(X,Y) from point A(0,0) with initial speed v. It is possible with two launch angles, Φ₁ and Φ₂, the times of flight are t₁ and t₂ respectively.

Problem 1) says that the line (A,B) encloses the angle θ with the horizontal (x axis), that is, Y/X=tan(θ). Find if the expression

$E=t_1^2+t_2^2+2t_1 t_2\sin(\theta)$

depends on θ.

Problem 2) says that the initial speed is v=√(2gh), and the trajectories cross each other at 90° angle at B. Show that B is on an ellipse and find out which it is from the given ones.

If a projectile is aimed to hit a point (X,Y) the tangent of the possible launch angles are

$\tan(\phi)=\frac{v^2 \pm \sqrt{v^4-g^2 X^2-2v^2 gY}}{gx}$

With the notation $C=v^4-g^2 X^2-2v^2 gY$,
the tangent of the angles are
$$\tan(\phi_1)=\frac{v^2 + \sqrt{C}}{gx}$$
and
$$\tan(\phi_2)=\frac{v^2 - \sqrt{C}}{gx}$$

The time of flight from A to B is is
$$t=\frac{X}{v \cos(\phi)}$$.
With that, the expression E is
$$E=t_1^2+t_2^2+2t_1 t_2\sin(\theta)=\frac{X^2}{v^2 \cos^2(\phi_1)}+\frac{X^2}{v^2 \cos^2(\phi_2)}+2\frac{X}{v \cos(\phi_1)}\frac{X}{v \cos(\phi_2)}\sin(\theta)$$

Using the relation
$$\frac{1}{cos^2(\phi)}=1+\tan^2(\phi)$$

$$E=\frac{X^2}{v^2}\left(1+\tan^2(\phi_1)+1+\tan^2 ( \phi_2)+2\sin(\theta) \sqrt{\left(1+\tan^2(\phi_1)\right) \left(1+\tan^2(\phi_2)\right)} \right)$$
Plug in the expressions for tan(Φ₁) and tan(Φ₂), expand and simplify.

In problem 2), start with the equation of the trajectory:
$$Y=\tan(\phi) X-\frac{gX^2}{2 v^2 \cos^2(\phi)}$$.
the tangent of the trajectory is :
$$Y'=\tan(\phi) -\frac{gX}{v^2 \cos^2(\phi)}$$.
If the two trajectories are perpendicular at B, the product of their tangents is -1.

$$\left( \tan(\phi_1) -\frac{gX}{v^2 \cos^2(\phi_1)} \right) \left ( \tan(\phi_2) -\frac{gX}{v^2 \cos^2(\phi_2)} \right) =-1$$.

Substitute 1+tan²(Φ) for 1/cos²(Φ), and also the expressions for tan(Φ₁) and tan(Φ₂), expand and simplify.

ehild

 

[sankalpmittal]

I show the solution steps a bit further:
The problem: you want to hit a point B(X,Y) from point A(0,0) with initial speed v. It is possible with two launch angles, Φ₁ and Φ₂, the times of flight are t₁ and t₂ respectively.

Problem 1) says that the line (A,B) encloses the angle θ with the horizontal (x axis), that is, Y/X=tan(θ). Find if the expression

$E=t_1^2+t_2^2+2t_1 t_2\sin(\theta)$

depends on θ.

Problem 2) says that the initial speed is v=√(2gh), and the trajectories cross each other at 90° angle at B. Show that B is on an ellipse and find out which it is from the given ones.

If a projectile is aimed to hit a point (X,Y) the tangent of the possible launch angles are

$\tan(\phi)=\frac{v^2 \pm \sqrt{v^4-g^2 X^2-2v^2 gY}}{gx}$

With the notation $C=v^4-g^2 X^2-2v^2 gY$,
the tangent of the angles are
$$\tan(\phi_1)=\frac{v^2 + \sqrt{C}}{gx}$$
and
$$\tan(\phi_2)=\frac{v^2 - \sqrt{C}}{gx}$$

The time of flight from A to B is is
$$t=\frac{X}{v \cos(\phi)}$$.
With that, the expression E is
$$E=t_1^2+t_2^2+2t_1 t_2\sin(\theta)=\frac{X^2}{v^2 \cos^2(\phi_1)}+\frac{X^2}{v^2 \cos^2(\phi_2)}+2\frac{X}{v \cos(\phi_1)}\frac{X}{v \cos(\phi_2)}\sin(\theta)$$

Using the relation
$$\frac{1}{cos^2(\phi)}=1+\tan^2(\phi)$$

$$E=\frac{X^2}{v^2}\left(1+\tan^2(\phi_1)+1+\tan^2 ( \phi_2)+2\sin(\theta) \sqrt{\left(1+\tan^2(\phi_1)\right) \left(1+\tan^2(\phi_2)\right)} \right)$$
Plug in the expressions for tan(Φ₁) and tan(Φ₂), expand and simplify.

But sinθ still remains. Answer says that this expression t_1^2+t_2^2+2t_1 t_2\sin(\theta) does not depend on θ.

What I want to say is this :
In the below image , one can easily see that this projectile is not symmetric. But arbitrarily assigning line AB as x-axis , one can convert it into a symmetric projectile.

http://postimage.org/image/ptms5vrwh/

In problem 2), start with the equation of the trajectory:
$$Y=\tan(\phi) X-\frac{gX^2}{2 v^2 \cos^2(\phi)}$$.
the tangent of the trajectory is :
$$Y'=\tan(\phi) -\frac{gX}{v^2 \cos^2(\phi)}$$.
If the two trajectories are perpendicular at B, the product of their tangents is -1.

$$\left( \tan(\phi_1) -\frac{gX}{v^2 \cos^2(\phi_1)} \right) \left ( \tan(\phi_2) -\frac{gX}{v^2 \cos^2(\phi_2)} \right) =-1$$.

Substitute 1+tan²(Φ) for 1/cos²(Φ), and also the expressions for tan(Φ₁) and tan(Φ₂), expand and simplify.

ehild

Thanks ! I'll at once do it !

 

[ehild]

But sinθ still remains. Answer says that this expression t_1^2+t_2^2+2t_1 t_2\sin(\theta) does not depend on θ.

Both t1 and t2 contain theta, and sin theta cancels with them. Try.

What I want to say is this :
In the below image , one can easily see that this projectile is not symmetric. But arbitrarily assigning line AB as x-axis , one can convert it into a symmetric projectile.

If you change the direction of gravity, too, perpendicular to AB. Have you got such a great power? :rofl:

ehild

 

[sankalpmittal]

Needs to be deleted. Unintentional double posting.

 

[sankalpmittal]

Both t1 and t2 contain theta, and sin theta cancels with them. Try.

If you change the direction of gravity, too, perpendicular to AB. Have you got such a great power? :rofl:

ehild

Ah !

So resolving g as gcosθ perpendicular to AB and gsinθ (no use till now) , I get ,

t₁ = 2usin(β-θ)/gcos(θ)
t₂ = 2usin(90-(β-θ))/gcos(θ)
= 2ucos(β-θ)/gcos(θ)

If I resolved components of g perpendicular to AB and made this projectile symmetric and range of two projectiles are same , the two angles will be complementary.

So I got two equations for t₁ and t₂. Now I tried evaluating
E=t₁²+t₂²+2t₁t₂sin(θ) from here but θ still remained.

Am I going correct. You say to plug tan values in terms of x , y and all ? Theta is not canceling up.

Problem 2 is over. No need to worry about it now.

 

[ehild]

So resolving g as gcosθ perpendicular to AB and gsinθ (no use till now) , I get ,

t₁ = 2usin(β-θ)/gcos(θ)
t₂ = 2usin(90-(β-θ))/gcos(θ)
= 2ucos(β-θ)/gcos(θ)

If I resolved components of g perpendicular to AB and made this projectile symmetric and range of two projectiles are same , the two angles will be complementary.

There are two angles to reach a point (x,y) from the origin, but the angles are complementary only when the initial and final heights are the same.
In that case, the projectile flies upward for time vsin(β)/g, and back for the same time. So the time of flight is t=2vsin(β)/g.
The projectile moves with constant velocity v_x=vcos(β) in the horizontal direction, so x=vcos(β)2vsin(β)/g=v²/g sin(2β). If you know that the range of the projectile must be x, the launch angle is obtained from the equation sin(2β)=gx/v². There are two solutions for 2β: 2β₁=arcsin(gx/v²) and 2β₂=pi-arcsin(gx/v²). That means β₂=pi/2-β₁.

If the final height is not the same as the initial height, the time of flight is not 2vsinβ/g, so the above argument can not be applied, the angles are not complimentary. This does not change if you choose a different system of coordinates.

In your system of coordinates g has both y and x components. The projectile accelerates in the x direction so x≠vcos(β-θ)2vsin(β-θ)/(gcosθ). The angles are not complimentary.

Imagine θ=80°. Then you need to launch both projectiles at a greater angle but still less than 90°. So both launch angles are less than 10°. Their sum can not be 90°.

ehild

 

[sankalpmittal]

There are two angles to reach a point (x,y) from the origin, but the angles are complementary only when the initial and final heights are the same.
In that case, the projectile flies upward for time vsin(β)/g, and back for the same time. So the time of flight is t=2vsin(β)/g.
The projectile moves with constant velocity v_x=vcos(β) in the horizontal direction, so x=vcos(β)2vsin(β)/g=v²/g sin(2β). If you know that the range of the projectile must be x, the launch angle is obtained from the equation sin(2β)=gx/v². There are two solutions for 2β: 2β₁=arcsin(gx/v²) and 2β₂=pi-arcsin(gx/v²). That means β₂=pi/2-β₁.

If the final height is not the same as the initial height, the time of flight is not 2vsinβ/g, so the above argument can not be applied, the angles are not complimentary. This does not change if you choose a different system of coordinates.

In your system of coordinates g has both y and x components. The projectile accelerates in the x direction so x≠vcos(β-θ)2vsin(β-θ)/(gcosθ). The angles are not complimentary.

Imagine θ=80°. Then you need to launch both projectiles at a greater angle but still less than 90°. So both launch angles are less than 10°. Their sum can not be 90°.

ehild

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 ?

 

[ehild]

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

 

[sankalpmittal]

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 ?

 

[ehild]

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

ehild

 

[sankalpmittal]

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

ehild

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,

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

The times of flight are

$$t_1=\frac{X}{v\cos(\phi_1)}$$
and
$$t_2=\frac{X}{v\cos(\phi_2)}$$.

Proceed.

ehild

 

[sankalpmittal]

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 ! :(

 

[ehild]

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

 

[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]

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.

 

[ehild]

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).
$$2 t_1 t_2 \sin(\theta)=\frac {4 \sin(\theta)}{g^2}\sqrt{(v^2-gY+C)(v^2-gY-C)}=\frac {4 \sin(\theta)}{g^2}\sqrt{(v^4-2gYv^2+g^2Y^2-C^2)}$$

 

[sankalpmittal]

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 got to 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 !)