AP EAMCET · PHYSICS · Motion In Two Dimensions
A body is projected with a speed \(u\) at an angle \(\theta\) with the horizontal. The radius of curvature of the trajectory, when it makes an angle \(\left(\frac{\theta}{2}\right)\) with the horizontal is ( \(g\)-acceleration due to gravity)
- A \(\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{\sqrt{3} g}\)
- B \(\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{2 g}\)
- C \(\frac{2 u^2 \cos ^3 \theta \sec ^2\left(\frac{\theta}{2}\right)}{g}\)
- D \(\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{g}\)
Answer & Solution
Correct Answer
(D) \(\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{g}\)
Step-by-step Solution
Detailed explanation
Let velocity of projectile is \(v\) at an angle \(\frac{\theta}{2}\) with horizontal \[ \begin{aligned} \therefore & v \cos \frac{\theta}{2}=u \cos \theta \\ \text { or } \quad v & =\frac{u \cos \theta}{\cos \frac{\theta}{2}} \end{aligned} \] As horizontal component remains…
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