TS EAMCET · Physics · Motion In Two Dimensions
It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point \(p\) at a horizontal distance \(r\) from the point of projection. If \(t_1\) and \(t_2\) are times taken to reach this point in two possible ways, then the product \(t_1 t_2\) is proportional to
- A \(\frac{1}{r}\)
- B \(r\)
- C \(r^2\)
- D \(\frac{1}{r^2}\)
Answer & Solution
Correct Answer
(B) \(r\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & T_1=\frac{2 u \sin \theta}{g} \\ & \text { and } \quad T_2=\frac{2 u \cos \theta}{g} \\ & T_1 T_2=\frac{2 u^2 \sin \theta \cos \theta}{g}=\frac{2 r}{g} \\ & {\left[\because R=\frac{v^2 \sin 2 \theta}{g}\right]} \\ & \end{aligned}\)
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