Question

A particle is projected from a point $O$ with an initial speed of $30m/s$ to pass through a point which is $40m$ from $O$ horizontally and $10m$ above. There are two angles of projection with the horizontal for which this is possible. If these angles are $\alpha$ and $\beta$, then find the value of $\tan [-(\alpha +\beta \left)\right]$

• A. $4$
• B. $2$
• C. $3$
• D. $1$

Answer 1

The correct option is A $4$

$y=x\tan \theta -\frac{1}{2}\frac{g{x}^2}{u^2}(1+{\tan }^2\theta )$

$y=x\tan \theta -\frac{x^2}{180}(1+{\tan }^2\theta )$

$10=40\tan \theta -\frac{80}{9}(1+ta{n}^2\theta )$

$8{\tan }^2\theta -36\tan \theta +17=0$
The equation is quadratic, so two values of $\theta$ can be assumed as $\alpha$ & $\beta$.
$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

While $\tan \alpha +tan\beta =\frac{9}{2}$ (by sum of two roots)

$\tan \alpha \tan \beta =\frac{17}{8}$
$\tan (\alpha +\beta )=-4$
$\tan (-[\alpha +\beta \left]\right)=4$