question_answer

  A body is projected at t = 0 with a velocity \[10\,m\,{{s}^{-1}}\] at an angle of \[60{}^\circ \] 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 \[g=10\,m\,{{s}^{-2}},\]the value of R is                 [JEE Main Online Paper (Held On 11-Jan-2019 Morning]            

A) 2.5 m                           

B) 2.8 m             

C) 10.3 m           

D)                  5.1 m     

Correct Answer: B

Solution :

At any time t, the horizontal velocity\[{{v}_{x}}=u\cos \theta \] Vertical velocity,\[{{v}_{y}}=(u\cos \theta )\] At \[t=1s,{{v}_{x}}=10\cos {{60}^{o}}=5m/s\]             \[{{v}_{y}}=(5\sqrt{3}-10)m/s\] The radius of curvature at time t, \[R=\frac{{{v}^{2}}}{a}=\frac{{{v}^{2}}}{g\cos \phi }=\frac{v_{x}^{2}+v_{y}^{2}}{g\cos \phi }\] \[=\frac{(25)+(75+100-100\sqrt{3})}{10\cos \phi }\] \[\tan \phi =\left| \frac{{{v}_{y}}}{{{v}_{x}}} \right|=\left| \frac{5\sqrt{3}-10}{5} \right|\]\[\Rightarrow \phi ={{15}^{o}}\] \[\therefore \]\[R=\frac{2.7}{\cos 15}=2.79\simeq 2.8m\]