Finding two variables in projectile motion

[Makonia]

Homework Statement

You are the chief of the Angry Birds (Google it, if you don’t know the reference...it’s not important). You can shoot red birds from a catapult, and they will fly as projectiles under the effect of gravity. Gravity points downwards and has magnitude g = 9.80 m/s2 . Your goal is to hit some nasty green pigs, who have stolen your Angry Bird Eggs. You are able to adjust the angle θ and speed v of the projectile, as it leaves the catapult. The projectile leaves the catapult at a height h above the ground.

a) First, consider the case when a single green pig is placed on the ground, at a distance d from the foot of the catapult. Find a relation between the angle and speed required to hit it. Remember to draw a sketch of the situation.

b) Next consider the case where there are two green pigs. One on the ground at a distance d as before, the other at a distance l < d, and placed on a pole of height h. What should you choose for θ and v in order to hit both pigs 1 with the same red bird? Why must one require l < d? Remember to draw a sketch of the situation.

Homework Equations

This task is related to projectile motion and I'm therefor pretty sure I'm supposed to solve it with the projectile motion equations.
In other words x = Vx(initial)*t because there is no acceleration in the x-axis direction and y = Vy(initial)t - 1/2gt^2
This is by the way a question i got at university but I'm posting it here because the physics seen rather basic, the problem is rather that in task b i need to determine two variabels and I'm not given the distance. My point is that if calculus or som other more advanced methods like advanced algebra could be used to solve this then that should not be a problem for me.
g = gravity = 9.8 m/s^2
d is the distance between 0 and the last pig
θ = is the angle the projectile was shot out with

The Attempt at a Solution

In task a I found an expression for t in x-direction by using the x = Vxt (and setting x to d) and got t = d/Vx
I put this expression for t into the equation in y-dirextion: y = Vyt - 1/2gt^2. By setting y to -h because i think like the projectile is shot from h which is 0 and land on the ground, -h. Also I've used that Vx = Vcosθ an Vy = Vsinθ
I solved this equation for V and as result I got:
V = sqrt((g*d^2)/(2*cos^2(θ)(h+d*tanθ)))
This is the answer I think is correct in task a and I think I might have to use it to solve b but I am not sure.
The b task is difficult because I need to find both θ and the innitial speed in order to hit both pigs an by my logic there is an endles different possible answers to that.
I've been trying to find equations for t or v but I have no idea how to make sure it hits both targets and write an equation on that so if anyone think they can solve it or help me by giving me som hints or pointers to where i should start at least that would be most appreciated. Also sorry for my bad english but I hope you understand

 

[TSny]

Your result for part (a) looks good to me.

For part (b), try deriving a similar relation between V and θ for hitting the bird on the pole. Then you will have two equations for the two unknowns V and θ.

 

[Makonia]

Your result for part (a) looks good to me.

For part (b), try deriving a similar relation between V and θ for hitting the bird on the pole. Then you will have two equations for the two unknowns V and θ.

I tried to do the same thing as in a but replacing d with l instead and setting h to 0 because the projectile is shot and is landing in the same height so therefor I put it to 0
Got this equation: 0=l*tanθ - g/2*(l^2)/(v^2cos^2(θ)). Should I also solve this equation for v you think or how should I find v and θ? θ seems especially difficult to me because there's both tanθ and cos^2(θ)

 

[TSny]

Got this equation: 0=l*tanθ - g/2*(l^2)/(v^2cos^2(θ))

Looks good. You might try rearranging this to get g/2*(l^2)/(v^2cos^2(θ)) on one side.
Similarly, from your relation in part (a), solve for g/2*(d^2)/(v^2cos^2(θ)) .

See what happens if you then divide the two equations.

 

[Makonia]

Looks good. You might try rearranging this to get g/2*(l^2)/(v^2cos^2(θ)) on one side.
Similarly, from your relation in part (a), solve for g/2*(d^2)/(v^2cos^2(θ)) .

See what happens if you then divide the two equations.

This is the two equations I then get. I've never devided two equations on each other so not quite sure how to do it but depending on whitch of them i divide on the other I will either get l/d + h or d/l + h on the right side. The left I don't know, it seems to me I get l^2/d^2 or the other way around. Was this what you ment by divide the two equations?

 

[jensjensen]

You can put tanθ/l=tanθ/d+h/d^2 and solve for θ

 

[TSny]

This is the two equations I then get. I've never devided two equations on each other so not quite sure how to do it but depending on whitch of them i divide on the other I will either get l/d + h or d/l + h on the right side. The left I don't know, it seems to me I get l^2/d^2 or the other way around. Was this what you ment by divide the two equations?

Yes. Or you can divide the top equation by $l^2$ and the bottom equation by $d^2$. Then the left hand sides are equal.

 

[tom ryen]

Is the answer θ=tan^-1(-hl/ld-d^2) ?

 

[TSny]

Is the answer θ=tan^-1(-hl/ld-d^2) ?

EDIT: Well, I don't want to say if this is the correct answer because Makonia has not yet posted an answer.

EDIT 2: Now that Makonia has given her answer, I can comment that I think your answer is correct. By the way, welcome to PF!

 

[jensjensen]

I got the same thing. But not sure if it is correct, nor the best answer.

 

[Makonia]

EDIT: Well, I don't want to say if this is the correct answer because Makonia has not yet posted an answer.

I got θ=tan^-1(lh/d(d-l)) so I think it's correct

 

[TSny]

I got θ=tan^-1(lh/d(d-l)) so I think it's correct

OK. Great. I think that's the answer, too.

 

[hefalomp]

Yes. Or you can divide the top equation by $l^2$ and the bottom equation by $d^2$. Then the left hand sides are equal.

Yes, but then the right sides will be: tan(θ) / l) i the first eq. and in the second: (tan(θ) + h ) / d, and when you put these two equal to one another the tan will go away?
So how did you then get θ=tan^-1(lh/d(d-l)) ?

 

[TSny]

Yes, but then the right sides will be: tan(θ) / l) i the first eq. and in the second: (tan(θ) + h ) / d, and when you put these two equal to one another the tan will go away?
So how did you then get θ=tan^-1(lh/d(d-l)) ?

The tan will not go away due to the fact that tan on one side is divided by l and tan on the other side is divided by d.

 

[klippen88]

Hi I'm only getting tan^-1( lh/d(1-l ), i don't understand how you guys find "d^2'" in the nominator, please help...

 

[TSny]

Hi I'm only getting tan^-1( lh/d-dl ), i don't understand how you guys find two "d's" in the nominator, please help...

Welcome to PF!

I believe you are saying that your denominator is d - dl. But note that it doesn't make sense to subtract these two terms. d had the dimensions of length while dl has the dimensions of length squared. If you show us your individual steps, we can track down your source of error.

As a sort of nit-picky point, parentheses should be used around the denominator as so: tan^-1( lh/(d-dl )). Otherwise, it would be interpreted as dividing lh by just d rather than (d - dl). Also, there are formatting icons for superscripts, so you can write it as tan^-1[ lh/(d-dl )].

 

[TSny]

I got the same thing. But not sure if it is correct, nor the best answer.

Welcome to PF! It's good to see that we are all getting the same result.

 

[klippen88]

I believe you are saying that your denominator is d - dl. But note that it doesn't make sense to subtract these two terms. d had the dimensions of length while dl has the dimensions of length squared. If you show us your individual steps, we can track down your source of error.

As a sort of nit-picky point, parentheses should be used around the denominator as so: tan^-1( lh/(d-dl )). Otherwise, it would be interpreted as dividing lh by just d rather than (d - dl). Also, there are formatting icons for superscripts, so you can write it as tan^-1[ lh/(d-dl )].

I'm sorry I wrote my answer in the stress and chaos of all this. My answer is of course tan^-1( lh/d(1-l )) (NOT "d-dl" in denominator). This will be length vs. length, so the dimensions adds up. But if i put in imaginary numbers for l, h & d. I'm getting a negativ angel, so my calculation must be wrong.
I'll show you how I have done it:
tanΘ/l = tanΘ+h/d
I multiply with l
and move l*tanΘ to the other side.
tanΘ-l*tanΘ=lh/d
tanΘ(1-l)=lh/d
Tan^-1(lh/d(1-l))

 

[jensjensen]

I'm sorry I wrote down my answer wrong in the stress and chaos of all this. My answer is of course tan^-1( lh/d(1-l )) (NOT "d-dl" in denominator). This will be length vs. length, so the dimensions adds up. But if i put in imaginary numbers for l, h & d. I'm getting a negativ angel, so my calculation must be wrong.
I'll show you have I have done it:
tanΘ/l = tanΘ+h/d
I multiply with l
and move l*tanΘ to the other side.
tanΘ-l*tanΘ=lh/d
tanΘ(1-l)=lh/d
Tan^-1(lh/d(1-l))

You have tanΘ/l = tanΘ/d+h/d^2
Then you end up with tan^-1(lh/d(d-l))=Θ

 

[klippen88]

You have tanΘ/l = tanΘ/d+h/d^2
Then you end up with tan^-1(lh/d(d-l))=Θ

oooh that's why, thnx man :-D

 

[hefalomp]

I'm sorry I wrote my answer in the stress and chaos of all this. My answer is of course tan^-1( lh/d(1-l )) (NOT "d-dl" in denominator). This will be length vs. length, so the dimensions adds up. But if i put in imaginary numbers for l, h & d. I'm getting a negativ angel, so my calculation must be wrong.
I'll show you how I have done it:
tanΘ/l = tanΘ+h/d
I multiply with l
and move l*tanΘ to the other side.
tanΘ-l*tanΘ=lh/d
tanΘ(1-l)=lh/d
Tan^-1(lh/d(1-l))

I have done the same as you is this the right answear? or should it be tan^-1(lh/d(d-l))=Θ, if so what did you change?

 

[jensjensen]

you have tan(θ)/l=tanθ/d+h/d^2
Which is the same as tanθ(1/l-1/d)=h/d^2
You can write that as tanθ=hl/(d(d-l)) ->θ=tan^-1(hl/(d(d-l)))

 

[hefalomp]

you have tan(θ)/l=tanθ/d+h/d^2
Which is the same as tanθ(1/l-1/d)=h/d^2
You can write that as tanθ=hl/(d(d-l)) ->θ=tan^-1(hl/(d(d-l)))

I got tan(θ)/l=(tanθ+h)/d and not tan(θ)/l=tanθ/d+h/d^2, how did you get tan(θ)/d+h/d^2

 

[Fjompen11]

Hi, I have got the same equation for the angle, but I am not sure how to find the equation for V. help please.