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.
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}} -...
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)...
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...