Question

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

• 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

Answer: (C)

${\tan }^{-1}\frac{7}{4}$

From question,
Horizontal velocity (initial),
$u_x=\frac{40}{2}=20m/s$
Vertical velocity (initial), $50=u_{y}t+\frac{1}{2}g{t}^2$
$\Rightarrow u_y\times 2+\frac{1}{2}(-10)\times 4$
or, $50=2u_y-20$
or, $u_y=\frac{70}{2}=35m/s$
$\therefore \tan \theta =\frac{u_y}{u_x}=\frac{35}{20}=\frac{7}{4}$
$\Rightarrow$ Angle $\theta ={\tan }^{-1}\frac{7}{4}$.