A particle is projected at speed $u$, at angle $\alpha$, from level ground. $\theta$ is the angle of elevation of the projectile from the initial point of projection and $\phi$ is the angle that the velocity of the projectile makes with the horizontal. Show that $\tan\alpha+\tan\phi=2\tan\theta$ at all points on the trajectory.

My attempt:

The equations for the motion of the projectile are $$\ddot{x}=0, \ddot{y}=-g, \dot{x}=u\cos\alpha, \dot{y}=u\sin\alpha-gt, x=tu\cos\alpha, y=tu\sin\alpha-\dfrac{1}{2}gt^2$$

$$\begin{aligned}\tan\theta&=\dfrac{y}{x}=\tan\alpha-\dfrac{g}{2u}\sec\alpha \quad (\star)\\ \tan\phi&=\dfrac{\dot{y}}{\dot{x}}=\tan\alpha-\frac{gt}{2u}\sec\alpha\\ \end{aligned}$$

Rearranging the second equation and substituting into $(\star)$, we get $$\tan\theta=\tan\alpha-\dfrac{\tan\alpha-\tan\phi}{t}$$ or $$t\tan\theta=(t-1)\tan\alpha+\tan\phi$$ Now substituting $t=2$ gives $$2\tan\theta=\tan\alpha+\tan\phi.$$ But this does not seem quite right - I have only showed it's true when $t=2$. Is my approach wrong?


1 Answer 1


We have

$$ \tan\phi=\dfrac{\dot{y}}{\dot{x}}=\frac{u\sin \alpha-gt}{u\cos \alpha}=\tan\alpha-\frac{gt}{u}\sec\alpha $$

(here you have a $2$ at the denominator that is wrong)

and from here: $$ \frac{gt}{u}\sec \alpha=\tan \alpha -\tan \phi $$

that, substituted in $(\star)$ gives the correct result.


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