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

In summary: First, divide both sides by ΔX to get \frac{ΔY}{ΔX} = Tanθ + \frac{ A ΔX}{2V_0^2} (1 + Tan^2θ) Then, let u = Tanθ Which means \frac{du}{dθ} = Sec^2θ \frac{dθ}{dθ} = Sec^2θ And \frac{d}{dθ} (1 + Tan^2θ) = 2 Tanθ Sec^2θ So that the original equation becomes \frac{ΔY}{ΔX} = u + \frac{A ΔX}{2V_0
  • #1
Marven345
5
0

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



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

[itex] ΔY = -0.5 m [/itex]

[itex] Vx = V_0 Cos θ [/itex]

[itex] Vy = V_0 Sin θ [/itex]

[itex] ΔY = V_0 Sin θ T + \frac{AT^2}{2}[/itex]

[itex] ΔX = V_0 Cos θ T [/itex]

The Attempt at a Solution



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

[itex] T = \frac{ΔX}{V_0 Cos θ} [/itex]

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

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

Simplified

[itex] ΔY = \frac{ΔX Sin θ}{Cos θ} + \frac{AΔX^2}{2V_0^2 Cos^2 θ} [/itex]

Then I multiplied both sides by [itex]Cos^2 θ[/itex].

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

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.
 
Last edited:
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  • #2
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 ##.
 
  • #3
Thanks for the reply, though I don't see how this would work. Would I turn the

[itex] ΔYCos^2 θ [/itex] into

[itex] ΔY(1-Sin^2 θ) [/itex]

or

[itex] Δ Y - ΔY Sin^2 θ [/itex] ?

I'm not sure where this would lead me.

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

[itex] Sinθ [/itex]

in the [itex] (SinθCosθ ΔX) [/itex] part, so I can turn into a quadratic equation, but I don't know how that would work.
 
  • #4
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.
 
  • #5
Oh, so I would use

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

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

Set it equal to 0

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

and try to use quadratics.

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

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.
 
  • #6
Marven345 said:
Oh, so I would use

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

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

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.
 
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  • #7
So
[itex] ΔYCos^2θ- \frac{AΔX^2}{2V_0^2} = \sqrt{1-Cos^2θ}CosθΔX [/itex]

Square both sides

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

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

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

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

Bring [itex]Cos^2θΔX^2[/itex] to the right as well

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

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

Where [itex](Cos^2θ)[/itex] = a very horrendous equation.

And square root it.

Thank you so much!
 
Last edited:
  • #8
Agreed. Not something to be enjoyed.

How about using ## \tan \theta ## instead, in the penultimate equation in #1?
 
  • #9
Oh, so would I use
[itex] Cos^2θ = \frac{1}{Sec^2θ} = \frac{1}{1+ Tan^2θ} [/itex]

Which equals

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

And use more quadratics

My head hurts x.x

Thank you for guiding me through.
 
  • #10
I think this one will be much easier on your head :)
 

1. What is a projectile problem and how is it solved?

A projectile problem is a physics problem that involves calculating the trajectory of an object launched into the air. It is solved using the equations of motion and the initial conditions of the object, such as its initial velocity and angle of launch.

2. What is the significance of solving for theta in a projectile problem?

Solving for theta, or the angle of launch, is essential in determining the initial direction of the projectile. It also affects the maximum height and range of the projectile.

3. How do you calculate theta given delta x, delta y, and V0 in a projectile problem?

To calculate theta, you can use the equation theta = tan^-1(delta y / delta x) - 1/2 * g * (delta x / V0^2), where g is the acceleration due to gravity and V0 is the initial velocity of the projectile.

4. Are there any other factors that need to be considered when solving for theta in a projectile problem?

Yes, other factors such as air resistance, wind speed, and surface conditions can also affect the trajectory of a projectile and should be taken into account when solving for theta.

5. Can theta have multiple solutions in a projectile problem?

Yes, in some cases, there can be multiple solutions for theta in a projectile problem. This can occur when the object reaches the same height or distance at different angles of launch. In these cases, the angle with the greatest initial velocity is usually the most accurate and preferred solution.

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