If the velocity of projection is 4 m/sec and the angle of projection is $${\alpha ^ \circ }$$, the maximum height of the projectile from a horizontal plane, is
A. $$\frac{{{{\text{u}}^2}{{\cos }^2}\alpha }}{{2{\text{g}}}}$$
B. $$\frac{{{{\text{u}}^2}{{\sin }^2}\alpha }}{{2{\text{g}}}}$$
C. $$\frac{{{{\text{u}}^2}{{\tan }^2}\alpha }}{{2{\text{g}}}}$$
D. $$\frac{{{{\text{u}}^2}\sin2\alpha }}{{2{\text{g}}}}$$
Answer: Option B

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