Question

The position of a projectile launched from the origin at $t=0$ is given by $\overrightarrow{r}=(40\hat{i}+50\hat{j})\text{m}$ at $t=2\text{s}$. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is
$(\text{Take}g=10{\text{ms}}^{-2})$

• A. ${\tan }^{-1}\frac{2}{3}$
• B. ${\tan }^{-1}\frac{3}{2}$
• C. ${\tan }^{-1}\frac{7}{4}$
• D. ${\tan }^{-1}\frac{4}{5}$

Answer 1

The correct option is C ${\tan }^{-1}\frac{7}{4}$

As we know that horizontal component of projected velocity remains constant.

So, from question,
Horizontal velocity, $u_x=\frac{40-0}{2-0}=20\text{m/s}$

And, initial vertical velocity $\left(u_y\right)$,

$s_y=u_{y}t+\frac{1}{2}a{t}^2$

$50=u_y\left(2\right)-\frac{1}{2}\left(10\right)(2{)}^2$

$\Rightarrow u_y=\frac{70}{2}=35\text{m/}s$

$\therefore \tan \theta =\frac{u_y}{u_x}=\frac{35}{20}=\frac{7}{4}$

$\Rightarrow \text{Angle},\theta ={\tan }^{-1}\frac{7}{4}$

Hence, option $\left(\text{B}\right)$ is correct.