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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

  1. A \(\frac{v}{2} \sqrt{1+2 \cos ^2 \theta}\)
  2. B \(\frac{v}{2} \sqrt{1+2 \sin ^2 \theta}\)
  3. C \(\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}\)
  4. D \(v \cos \theta\)
Verified Solution

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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