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

## 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.

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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: $\cos^2 \theta + \sin^2 \theta = 1$.

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.

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.

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

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

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

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.

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.

• 1 person
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 $Cos^4θ$

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

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!

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Agreed. Not something to be enjoyed.

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

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

Which equals

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