Question

At what angle should a body be projected with a velocity $24{\text{ms}}^{–1}$ just to pass over the obstacle $14\text{m}$ high at a distance of $24\text{m}$.
[Take $g=10{\text{ms}}^{–2}$]

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

Answer 1

Answer: (A), (B)

$\tan \theta =1$

From relations of projectile,
$x=u\cos \theta .t=24$
$\Rightarrow t=\frac{24}{24}\cos \theta =\frac{1}{\cos \theta }$
$y=u\sin \theta t-\frac{1}{2}g{t}^2$
$\Rightarrow 14=\frac{u\sin \theta }{\cos \theta }-\frac{5}{{\cos }^2\theta }$
$\Rightarrow 14=\frac{u\sin \theta }{\cos \theta }-\frac{5}{{\cos }^2\theta }$
$\Rightarrow 14=u\tan \theta -5{\sec }^2\theta$
$14=24\tan \theta -5(1+{\tan }^2\theta )$
$\Rightarrow 5{\tan }^2\theta -24\tan \theta +19=0$
$\tan \theta =\frac{24\pm \sqrt{{24}^2-4\left(5\right)\left(19\right)}}{2\times 5}=\frac{24\pm 14}{10}$
$\tan \theta =\frac{24+14}{10}=3.8$
$\tan \theta =\frac{24-14}{10}=\frac{10}{10}=1$