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The position of a projectile launched from the origin at $t=0$ is given by $\vec{r}=(40 \hat{i}+50 \hat{j}) \mathrm{m}$ at $t=$ $2 \mathrm{~s}$. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is $\left(\right.$ take $\left.g=10 \mathrm{~ms}^{-2}\right)$\
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Verified Answer
The correct answer is:
$\tan ^{-1} \frac{7}{4}$
From question, Horizontal velocity (initial),
$$
u_{x}=\frac{40}{2}=20 \mathrm{~m} / \mathrm{s}
$$
Vertical velocity (initial), $50=u_{y} t+\frac{1}{2} g t^{2}$
$$
\begin{array}{l}
\Rightarrow u_{y} \times 2+\frac{1}{2}(-10) \times 4 \\
\text { or, } \quad 50=2 u_{y}-20
\end{array}
$$
or, $u_{y}=\frac{70}{2}=35 \mathrm{~m} / \mathrm{s}$
$$
\begin{array}{l}
\therefore \quad \tan \theta=\frac{u_{y}}{u_{x}}=\frac{35}{20}=\frac{7}{4} \\
\Rightarrow \quad \text { Angle } \theta=\tan ^{-1} \frac{7}{4}
\end{array}
$$
$$
u_{x}=\frac{40}{2}=20 \mathrm{~m} / \mathrm{s}
$$
Vertical velocity (initial), $50=u_{y} t+\frac{1}{2} g t^{2}$
$$
\begin{array}{l}
\Rightarrow u_{y} \times 2+\frac{1}{2}(-10) \times 4 \\
\text { or, } \quad 50=2 u_{y}-20
\end{array}
$$
or, $u_{y}=\frac{70}{2}=35 \mathrm{~m} / \mathrm{s}$
$$
\begin{array}{l}
\therefore \quad \tan \theta=\frac{u_{y}}{u_{x}}=\frac{35}{20}=\frac{7}{4} \\
\Rightarrow \quad \text { Angle } \theta=\tan ^{-1} \frac{7}{4}
\end{array}
$$
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