TS EAMCET · Physics · Motion In Two Dimensions
A body is projected from the ground at an angle of \(\tan ^{-1}(\sqrt{7})\) with the horizontal. At half of the maximum height, the speed of the body is ' \(n\) ' times the speed of projection. The value of ' \(n\) ' is
- A 2
- B \(\frac{1}{3}\)
- C \(\frac{4}{3}\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
For projectile motion, \(\begin{aligned} & \theta=\tan ^{-1}(\sqrt{7}) \Rightarrow \tan \theta=\sqrt{7} \\ & \text { At } y=\frac{H}{2}, v_y^2=u_y^2-2 g\left(\frac{H}{2}\right) \end{aligned}\)…
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