JEE Mains · Physics · STD 11 - 3.2 motion in plane
The position of a projectile launched from the origin at \(t = 0\) is given by \(\vec r = \left( {40\hat i + 50\hat j} \right)\,m\) 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 \({\tan ^{ - 1}}\frac{4}{5}\)
Answer & Solution
Correct Answer
(C) \({\tan ^{ - 1}}\frac{7}{4}\)
Step-by-step Solution
Detailed explanation
\begin{array}{l} \,\,\,\,\,\,\,\,\,From\,question,\\ \,\,\,\,\,\,\,\,\,Horizontal\,velocity\,\left( {initial} \right),\\ \,\,\,\,\,\,\,\,\,{u_x} = \frac{{40}}{2} = 20m/s\\ \,\,\,\,\,\,\,\,\,Vertical\,velocity\,\left( {initial} \right),\\ \,\,\,\,\,\,\,\,\,\,50 = {u_y}t +…
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