TS EAMCET · Physics · Motion In Two Dimensions
A projectile is fired at an angle of \(45^{\circ}\) with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is
- A \(60^{\circ}\)
- B \(\tan ^{-1}\left(\frac{1}{2}\right)\)
- C \(\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
- D \(45^{\circ}\)
Answer & Solution
Correct Answer
(B) \(\tan ^{-1}\left(\frac{1}{2}\right)\)
Step-by-step Solution
Detailed explanation
The given situation is shown in the following figure Angle of projection, \(\theta=45^{\circ}\) The elevation angle of the projectile at its highest point as seen from the point of projection is \(\alpha\). \(\therefore \text { In } \triangle P O M \text {, }\)…
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