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3-2.Motion in Plane
medium
The position of a projectile launched from the origin at $t=0$ is given by $\vec{r}=(40 \hat{i}+50 \hat{j}) m$ at $t=$ $2 s$. 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
$\tan ^{-1} \frac{4}{5}$
Solution
(c)
From question, Horizontal velocity (initial),
$u_x=\frac{40}{2}=20\,m / s$
Vertical velocity (initial), $50=u_y t+\frac{1}{2} g t^2$
$\Rightarrow u_y \times 2+\frac{1}{2}(-10) \times 4$
$\text { or, } 50=2 u_y-20$
$\text { or, } u_y=\frac{70}{2}=35\,m / s$
$\therefore \tan \theta=\frac{u_y}{u_x}=\frac{35}{20}=\frac{7}{4}$
$\Rightarrow \text { Angle } \theta=\tan ^{-1} \frac{7}{4}$
Standard 11
Physics
