TS EAMCET · Physics · Motion In Two Dimensions
A particle is projected from the ground with an initial speed of \(v\) at an angle of projection \(\theta\). The average velocity of the particle between its time of projection and time it reaches highest point of trajectory is
- A \(\frac{v}{2} \sqrt{1+2 \cos ^2 \theta}\)
- B \(\frac{v}{2} \sqrt{1+2 \sin ^2 \theta}\)
- C \(\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}\)
- D \(v \cos \theta\)
Answer & Solution
Correct Answer
(C) \(\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { We know, average velocity }=\frac{\text { displacement }}{\text { time }} \\ & \qquad v_{\text {av }}=\frac{\sqrt{\mu^2+R^2 / 4}}{T / 2}\end{aligned}\) where, \(H=\) maximum height \(=\frac{v^2 \sin ^2 \theta}{2 g}\) Range…
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