JEE Mains · Physics · STD 11 - 3.2 motion in plane
A body is projected at \(t = 0\) with a velocity \(10\,ms^{-1}\) at an angle of \(60^o\) with the horizontal. The radius of curvature of its trajectory at \(t=1\,s\) is \(R.\) Neglecting air resistance and taking acceleration due to gravity \(g = 10\,ms^{-2},\) the radius of \(R\) is ........ \(m\)
- A \(10.3\)
- B \(2.8\)
- C \(2.5\)
- D \(5.1\)
Answer & Solution
Correct Answer
(B) \(2.8\)
Step-by-step Solution
Detailed explanation
\begin{array}{l} at\,t = 1\\ \,{u_x} = 5,\,{u_y} = 5\sqrt 3 \\ {v_y} = 5\sqrt 3 - 10\,\,;\,\,\,{v_x} = 5\\ \tan \,\theta \, = - 2\left( {2 - \sqrt 3 } \right) \Rightarrow \theta = - {30^ \circ }\\ R = \frac{{{v^2}}}{{a \bot }} = \frac{{{{10}^2}}}{{\left( {10\,\cos \,{{30}^ \circ…
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