TS EAMCET · Physics · Motion In Two Dimensions
An object of mass \(2 m\) is projected with a speed of \(100 \mathrm{~ms}^{-1}\) at an angle \(\theta=\sin ^{-1}\left(\frac{3}{5}\right)\) to the horizontal. At the highest point, the object breaks into two pieces of same mass \(m\) and the first one comes to rest. The distance between the point of projection and the point of landing of the bigger piece (in metre) is (Given, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(3840\)
- B \(1280\)
- C \(1440\)
- D \(960\)
Answer & Solution
Correct Answer
(C) \(1440\)
Step-by-step Solution
Detailed explanation
Horizontal range of the object fired, \(R=\frac{u^2 \sin 2 \theta}{g}\) At the highest point, when object is exploded into two equal masses, then \(2 m u \cos \theta=m(0)+m v\) or \(v=2 u \cos \theta\) It means, the horizontal velocity becomes double at the highest point, hence…
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