Question 267 of 308

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

A${\text{tan}}^{-1}\dfrac{3}{2}$B${\text{tan}}^{-1}\dfrac{2}{3}$C${\text{tan}}^{-1}\dfrac{7}{4}$D${\text{tan}}^{-1}\dfrac{4}{5}$

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Answer: (C) ${\text{tan}}^{-1}\dfrac{7}{4}$

From question, Horizontal velocity (initial),
${u}_{x}=\dfrac{40}{2}=20m/s$
Vertical velocity (initial), $50={u}_{y}t+\dfrac{1}{2}g{t}^{2}$
$\Rightarrow 50={u}_{y}\times 2+\dfrac{1}{2}(-10)\times 4$
$50=2{u}_{y}-20$
${u}_{y}=\dfrac{70}{2}=35m/s$
$∴\tan \theta =\dfrac{{u}_{y}}{{u}_{x}}=\dfrac{35}{20}=\dfrac{7}{4}$
$\Rightarrow$ $\theta ={\tan }^{-1}\dfrac{7}{4}$