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3.2 , motion in plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics3.2 , motion in plane1 Mark
MCQ
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
    $\tan ^{-1} \frac{3}{2}$
  • ✓
    $\tan ^{-1} \frac{7}{4}$
  • D
    $\tan ^{-1} \frac{4}{5}$
✓

Answer

Correct option: C.
$\tan ^{-1} \frac{7}{4}$
c
(c)

From question, Horizontal velocity (initial),

$u_x=\frac{40}{2}=20\,m / s$

Vertical velocity (initial), $50=u_y t+\frac{1}{2} g t^2$

$\Rightarrow u_y \times 2+\frac{1}{2}(-10) \times 4$

$\text { or, } 50=2 u_y-20$

$\text { or, } u_y=\frac{70}{2}=35\,m / s$

$\therefore \tan \theta=\frac{u_y}{u_x}=\frac{35}{20}=\frac{7}{4}$

$\Rightarrow \text { Angle } \theta=\tan ^{-1} \frac{7}{4}$

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