Question

At what angle should a body be projected with a velocity $24\frac{m}{s}$ just to pass over the obstacle $14$m high of a distance of $24m$. (Take $g=10m/s^2$)

• A. $\tan \theta =3.8$
• B. $\tan \theta =1$
• C. $\tan \theta =3.2$
• D. $\tan \theta =2$

Answer 1

The correct option is A $\tan \theta =3.8$

Given : $u=24m/s$ $x=24m$ $y=14m$

Using, equation of trajectory $y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

Or $14=24\tan \theta -\frac{10\times {24}^{2}se{c}^2\theta }{2\times {24}^2}$

Or $14=24\tan \theta -5(1+{\tan }^2\theta )$ $(\because se{c}^2\theta =1+{\tan }^2\theta )$

Or $5{\tan }^2\theta -24\tan \theta +19=0$

Or $\tan \theta =\frac{24\pm \sqrt{{24}^2-4\left(5\right)\left(19\right)}}{2\times 5}=\frac{24\pm 14}{10}$

$⟹\tan \theta =\frac{24+14}{10}=3.8$