Projectile Problem: Solving Theta given Delta X, Delta Y and V0.

[Marven345]

Homework Statement

It's a projectile problem.

Basically, you launch a projectile starting from a height ΔY (0.5 m) at speed V0 at an angle θ. Then after time T, the projectile then hits the ground (0m) and sticks to the ground, traveling distance ΔX.

The question goes if you are given ΔX and V_0 (which are known variables, but there is no numerical number to go with it), solve for θ.

Essentially, solve for θ given Delta X, Delta Y, and V0

Homework Equations

Acceleration (A) = -9.81 m/s^2

ΔY = -0.5 m

Vx = V_0 Cos θ

Vy = V_0 Sin θ

ΔY = V_0 Sin θ T + \frac{AT^2}{2}

ΔX = V_0 Cos θ T

The Attempt at a Solution

I tried solving for T in terms of ΔX, V0, and θ by using the 6th equation.

T = \frac{ΔX}{V_0 Cos θ}

Then attempted to plug this value of T into the 5th equation.

ΔY = \frac{V_0ΔX Sinθ}{V_0Cosθ} + \frac{A}{2}\frac{ΔX^2}{V_0^2 Cos^2}

Simplified

ΔY = \frac{ΔX Sin θ}{Cos θ} + \frac{AΔX^2}{2V_0^2 Cos^2 θ}

Then I multiplied both sides by Cos^2 θ.

(ΔY Cos^2 θ) = (Sinθ Cos θ ΔX) + \frac{AΔX^2}{2V_0^2}

And now I'm stuck and I'm not sure what I should do. I don't know if my original approach will get me to my desired answer.

If any clarification is needed, I will gladly provide some.

Thank you, any help would be greatly appreciated.

 

[voko]

Your approach is correct in general. You are stuck because this stage requires a bit of trigonometry. See if the following equation is of use: $\cos^2 \theta + \sin^2 \theta = 1$.

 

[Marven345]

Thanks for the reply, though I don't see how this would work. Would I turn the

ΔYCos^2 θ into

ΔY(1-Sin^2 θ)

or

Δ Y - ΔY Sin^2 θ ?

I'm not sure where this would lead me.

My current plan is to think of a way to somehow eliminate the

Sinθ

in the (SinθCosθ ΔX) part, so I can turn into a quadratic equation, but I don't know how that would work.

 

[voko]

The equation I mentioned is a way to eliminate either the sine or the cosine, so you end up with just one trigonometric function throughout.

 

[Marven345]

Oh, so I would use

Sin θ = \sqrt{1-Cos^2θ} and plug it into the equation to get

ΔYCos^2θ= \sqrt{1-Cos^2θ}CosθΔX + \frac{AΔX^2}{2V_0^2}

Set it equal to 0

ΔYCos^2θ- \sqrt{1-Cos^2θ}CosθΔX - \frac{AΔX^2}{2V_0^2} = 0

and try to use quadratics.

Cosθ = \frac{ ΔX\sqrt{1-Cos^2θ}} - \sqrt{{1-Cos^2θ}^2ΔX^2 - 4(ΔY\frac{AΔX^2}{2V_0^2}}}{2ΔY}

For some reason its not converting into that fancy text.

And then try to solve that monster.

Wow this answer won't be pretty x.x

Thanks.

 

[voko]

Oh, so I would use

Sin θ = \sqrt{1-Cos^2θ} and plug it into the equation to get

ΔYCos^2θ= \sqrt{1-Cos^2θ}CosθΔX + \frac{AΔX^2}{2V_0^2}

Set it equal to 0

Nope. First you need to eliminate the radical. You can do that by collecting all the terms without the radical on one side of the equation, and then squaring both sides.

 

[Marven345]

So
ΔYCos^2θ- \frac{AΔX^2}{2V_0^2} = \sqrt{1-Cos^2θ}CosθΔX

Square both sides

ΔY^2Cos^4θ- \frac{2AΔX^2ΔYCos^2θ}{2V_0^2} + \frac{A^2ΔX^4}{4V_0^4} = (1-Cos^2θ) Cos^2θΔX^2

Which equals ΔY^2Cos^4θ- \frac{2AΔX^2ΔYCos^2θ}{2V_0^2} + \frac{A^2ΔX^4}{4V_0^4} = Cos^2θΔX^2 - Cos^4θΔX^2

Then I would bring the Cos^4θΔX^2 to the right side.ΔY^2Cos^4θ + Cos^4θΔX^2 - \frac{2AΔX^2ΔYCos^2θ}{2V_0^2} + \frac{A^2ΔX^4}{4V_0^4} = Cos^2θΔX^2

Factor out Cos^4θCos^4θ(ΔY^2 +ΔX^2) - \frac{2AΔX^2ΔYCos^2θ}{2V_0^2} + \frac{A^2ΔX^4}{4V_0^4} = Cos^2θΔX^2

Bring Cos^2θΔX^2 to the right as well

Cos^4θ(ΔY^2 +ΔX^2) - \frac{2AΔX^2ΔYCos^2θ}{2V_0^2} - Cos^2θΔX^2 + \frac{A^2ΔX^4}{4V_0^4} = 0

Factor it out, and use quadratics?Cos^4θ(ΔY^2 +ΔX^2) - (Cos^2θ)(\frac{2AΔX^2ΔY}{2V_0^2} - ΔX^2) + \frac{A^2ΔX^4}{4V_0^4} = 0

Where (Cos^2θ) = a very horrendous equation.

And square root it.

Thank you so much!

 

[voko]

Agreed. Not something to be enjoyed.

How about using $\tan \theta$ instead, in the penultimate equation in #1?

 

[Marven345]

Oh, so would I use
Cos^2θ = \frac{1}{Sec^2θ} = \frac{1}{1+ Tan^2θ}

Which equals

ΔY = ΔX Tanθ + \frac{ A ΔX^2(1 + Tan^2θ)}{2V_0^2}

And use more quadratics

My head hurts x.x

Thank you for guiding me through.

 

[voko]

I think this one will be much easier on your head :)