A) \[20\sqrt{3}\]
B) \[\frac{20}{\sqrt{3}}\]
C) 20
D) \[20\sqrt{2}\]
Correct Answer: C
Solution :
component of acceleration g of the projectile perpendicular to the velocity is radial acceleration Hence at Q \[g\cos {{30}^{0}}\,=\frac{{{v}^{2}}}{R}\] \[10\times \frac{\sqrt{3}}{2}\,=\frac{{{v}^{2}}3\sqrt{3}}{80}\] \[v=\frac{20}{\sqrt{3}}\,=(m/\sec )\] Now, \[u\cos {{60}^{0}}\,=v\,\cos {{30}^{0}}\] \[\frac{u}{2}\,=\frac{20}{\sqrt{3}}.\frac{\sqrt{3}}{2}\] \[u=20\,m/\sec \]You need to login to perform this action.
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