Question

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

• 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

Answer: (C)

$2u_{x}=40 \Rightarrow 4x=20$
$50=24y-\frac{1}{2}\times10\times2^{2}\Rightarrow 4y = 35$
$\tan\,\theta=\frac{u_{y}}{u_{x}}=\frac{35}{20}=\frac{7}{4}$
$\theta=\tan^{-1}\left(\frac{7}{4}\right)$