Question

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

• 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)

From question, Horizontal velocity (initial),
$u_{x}=\frac{40}{2}=20\,m/s$
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, $50=2 u_{y}-20$
or $u_{y}=\frac{70}{2}=35\, m/s$
$\therefore \tan \theta=\frac{u_{y}}{u_{x}}=\frac{35}{20}=\frac{7}{4}$
$\Rightarrow $ Angle $\theta=\tan ^{-1} \frac{7}{4}$