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Spunker 8
Lv 4
Spunker 8 asked in Science & MathematicsPhysics · 1 decade ago

Solve Physics Equation for Theta?

For some reason it cut off the end of it. Heres what i have so far:

Δy=Vi(sinθ)(Δx/Vi(cosθ))+.5(-9.81)(Δx/Vi(cosθ))^2

Update:

For some reason it cut off the end of it. Heres what i have so far:

Δy=Vi(sinθ)(Δx/Vi(cosθ))+.5(-9.81)(Δx/vi(cosθ))^2

Update 2:

I am given Δy and Δx and Vi but need to solve for θ. This is a hard one.

I have these 2 equations I chose to manipulate.

Δx=Vi(cosθ)t

Δy=Vi(sinθ)t+.5(-9.81)t^2

I solved the first equation for t and then plugged it into the second one when it asks for t. I am told it will be a quadratic equation when it gets simplified. Any help is greatly appreciated.

Δy=Vi(sinθ)(Δx/Vi(cosθ))+.5(-9.81)(Δx/Vi(cosθ))^2

Update 3:

For some reason it cut off the end of it. Heres what i have so far:

Δy=Vi(sinθ)(Δx/Vi(cosθ))+.5(-9.81)(Δx/Vi(cosθ))^2

1 Answer

Relevance
  • Doc W
    Lv 6
    1 decade ago
    Favorite Answer

    Hor

    s = ut gives x = VcosA. t where A is the angle

    t = x/VcosA

    Vertically

    s = y

    u = VsinA

    a = -g

    t = t = x/VcosA

    s = ut + 0.5at^2 gives

    y = VcosA(x/VsinA) - 0.5g(x/VcosA)^2

    y = xtanA - 0.5gx^2/(VcosA)^2

    y = xtanA - 0.5gx^2(secA)^2/V^2 is th equation of the path of the projectile.

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