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, \[R=\frac{{{v}^{2}}}{a}=\frac{{{v}^{2}}}{g\cos \phi }=\frac{v_{x}^{2}+v_{y}^{2}}{g\cos \phi }\] \[=\frac{(25)+(75+100-100\sqrt{3})}{10\cos \phi }\] \[\tan \phi =\left| \frac{{{v}_{y}}}{{{v}_{x}}} \right|=\left| \frac{5\sqrt{3}-10}{5} \right|\]\[\Rightarrow \phi ={{15}^{o}}\] \[\therefore \]\[R=\frac{2.7}{\cos 15}=2.79\simeq 2.8m\]You need to login to perform this action.
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