A) \[\frac{\sqrt{3}}{2}\frac{{{\text{v}}^{\text{2}}}}{\text{g}}\]
B) zero
C) \[\frac{{{\text{v}}^{\text{2}}}}{\sqrt{2}\text{g}}\]
D) \[\frac{\sqrt{3}}{16}\frac{{{\text{v}}^{\text{2}}}}{\text{g}}\]
Correct Answer: D
Solution :
[d] \[{{\text{V}}_{\text{h}}}\text{= V cos}\theta \] Where h is the maximum height \[{{V}_{h}}\times h=\left( v\cos \theta \right)\left( \frac{{{v}^{2}}{{\sin }^{2}}\theta }{2g} \right)\] \[=\frac{{{v}^{3}}{{\sin }^{2}}\theta \cos \theta }{2g}=\frac{\sqrt{3}\,{{v}^{3}}}{16g}\]You need to login to perform this action.
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