For any position A, show that $\tan\alpha + \tan\beta =\tan\theta$
MY SOLUTION:
We know that $\tan\theta=\frac{4h}{R}$
$$\tan\alpha=\frac{h}{x}$$ And $$\tan\beta = \frac{h}{R-x}$$ So $$\tan\alpha + \tan\beta = \frac{hR}{xR-x^2}$$ That’s as far as I could go up to. What should I do next?
On
You may continue by substituting the following results for a projectile into your expression
$$ x = v_0t\cos \theta $$ $$ R = \frac{v_0^2}{g}\sin2\theta $$ $$ h = v_0t\sin\theta - \frac{1}{2} g t^2 $$
and verify that
$$\frac{hR}{xR-x^2}= \tan\theta$$
The numerator is
$$hR = \left(x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta}\right)\frac{v_0^2\sin 2\theta}{g}=x\tan\theta \left(\frac{v_0^2\sin 2\theta}{g}-x \right)$$
and the denominator
$$x(R-x) = x\left(\frac{v_0^2\sin 2\theta}{g}-x \right)$$
Their ratio comes out as
$$\frac{hR}{xR-x^2}= \tan\alpha + \tan\beta= \tan\theta$$
Let's start with the equations of motions along the two axis. In the horizontal axis you have $$x=ut\cos\theta$$ In the vertical direction you have $$h=ut\sin\theta-\frac12 gt^2$$ Taking $t=\frac{x}{u\cos\theta}$ from the first equation and plugging it into the second you get $$h=x\tan\theta-\frac12 g\frac{x^2}{u^2\cos^2\theta}$$ First, to calculate $R$ you set $h=0$. One solution is obviously $x=0$. For the other solution $$0=\tan\theta-\frac12 g\frac{R}{u^2\cos^2\theta}$$ This yields $$R=2\frac{u^2}{g}\tan\theta\cos^2\theta$$ Now rewrite the formula for the height in terms of $R$ as $$h=x\tan\theta-\frac{x^2}R\tan\theta$$ From here it should be easy, just plug this into your $\tan\alpha+\tan\beta$ equation: $$\tan\alpha+\tan\beta=\frac{hR}{x(R-x)}=\frac{xR\tan\theta-x^2\tan\theta}{x(R-x)}=\tan\theta$$