A) \[\frac{2{{\text{v}}^{\text{2}}}\tan \theta }{\text{g}}\]
B) \[\frac{{{\text{v}}^{\text{2}}}\sec \theta }{\text{g}}\]
C) \[\frac{2{{\text{v}}^{\text{2}}}\tan \theta \sec \theta }{\text{g}}\]
D) \[\frac{{{\text{v}}^{\text{2}}}sin\theta }{\text{g}}\]
Correct Answer: C
Solution :
[c] If t is the time of flight, then \[0=vt-\frac{1}{2}\left( g\cos \theta \right){{t}^{2}}\Rightarrow t=\frac{2v}{g}g\cos \theta .\] \[R=0+\frac{1}{2}\left( g\sin \theta \right){{t}^{2}}=\frac{1}{2}g\sin \theta {{\left( \frac{2v}{g\cos \theta } \right)}^{2}}\]\[=\frac{2{{v}^{2}}}{g}\tan \theta \,\sec \theta \]
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