Question:medium

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

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Use position components at a given time to back-calculate velocity components.
Updated On: Mar 23, 2026
  • \(\tan^{-1}\!\left(\dfrac{2}{3}\right)\)
  • \(\tan^{-1}\!\left(\dfrac{3}{2}\right)\)
  • \(\tan^{-1}\!\left(\dfrac{7}{4}\right)\)
  • \(\tan^{-1}\!\left(\dfrac{4}{5}\right)\)
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The Correct Option is B

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