The position of a projectile launched from the origin at t = 0 is given by r→=(40i^+50j^)m at t=2s. If the projectile was launched at an angle θ from the horizontal, then θ is (take g = 10 ms-2)

r¯=40i^+50j^m at t=2secu cos⁡ θt=40 u sin⁡ θt−1/2gt2=50⇒u cos⁡ θ=20 u sin⁡ θ=35⇒u sin ⁡θu cos ⁡θ=3520 tan⁡=7/4⇒θ=tan−1⁡(7/4)

The position of a projectile launched from the origin at t = 0 is given by r→=(40i^+50j^)m at t=2s. If the projectile was launched at an angle θ from the horizontal, then θ is (take g = 10 ms-2)

r¯=40i^+50j^m at t=2secu cos⁡ θt=40 u sin⁡ θt−1/2gt2=50⇒u cos⁡ θ=20 u sin⁡ θ=35⇒u sin ⁡θu cos ⁡θ=3520 tan⁡=7/4⇒θ=tan−1⁡(7/4)

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

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$\tan^{- 1} \frac{2}{3}$

$\tan^{- 1} \frac{3}{2}$

$\tan^{- 1} \frac{7}{4}$

$\tan^{- 1} \frac{4}{5}$

answer is C.

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Detailed Solution

$\begin{array}{l} \bar{r} = 40 \hat{i} + 50 \hat{j} m at t = 2 \sec \\ u \cos θt = 40 u \sin θt - 1 / 2 {gt}^{2} = 50 \\ \Rightarrow u \cos θ = 20 u \sin θ = 35 \\ \Rightarrow \frac{u \sin θ}{u \cos θ} = \frac{35}{20} \tan = 7 / 4 \\ \Rightarrow θ = \tan^{- 1} (7 / 4) \end{array}$