Answer
The takeoff angle $\theta$ is $14^\circ$.
Work Step by Step
$$0.342D\cos\theta+h\cos^2\theta=\frac{16D^2}{V_0^2}$$
The question asks for $\theta$, which is the takeoff angle, the kind of angle of takeoff which you might encounter in action movie when a car starts to fly out of control after being hit.
That means the interval concerned here is $[0^\circ, 180^\circ)$. The car cannot fly at an angle smaller or larger than that.
1) Put in $V_0=60, D=80, h=2$ to the equation
$$0.342\times80\cos\theta+2\cos^2\theta=\frac{16\times80^2}{60^2}$$
$$27.36\cos\theta+2\cos^2\theta=\frac{102400}{3600}$$
$$27.36\cos\theta+2\cos^2\theta\approx28.44$$
2) Solve the equation for $\theta$:
$$27.36\cos\theta+2\cos^2\theta=28.44$$
$$2\cos^2\theta+27.36\cos\theta-28.44=0$$
$$\cos^2\theta+13.68\cos\theta-14.22=0$$
We take the equation as a quadratic formula, whose $a=1, b=13.68, c=-14.22$
- Find $\Delta$: $\Delta=b^2-4ac=13.68^2-4\times1\times(-14.22)=187.1424+56.88=244.0224$
- Find $\cos\theta$: $$\cos\theta=\frac{-b\pm\sqrt\Delta}{2a}=\frac{-13.68\pm\sqrt{244.0224}}{2}$$
- For $\cos\theta=\frac{-13.68+\sqrt{244.0224}}{2}\approx0.971$
We would have $$\theta=\cos^{-1}0.971\approx14^\circ\in[0^\circ,180^\circ)$$
This solution is accepted.
- For $\cos\theta=\frac{-13.68-\sqrt{244.0224}}{2}\approx-14.651$
As the range of a cosine function is $[-1,1]$, and $-14.651$ lies out of that range, there is no value of $\theta$ which $\cos\theta$ can equal $-14.651$
So overall, only one possibility of $\theta$ has been found. The takeoff angle $\theta$ is $14^\circ$.
