A body is projected with an initial velocity U at an angle O to the horizontal. The time taken by it to reach its maximum height is given by the expression
\(\frac{2u \text{sin}\theta}{g}\)
\(\frac{u^2 \text{sin}^2\theta}{2g}\)
\(\frac{u \text{sin}\theta}{g}\)
\(\frac{u \text{sin}\theta}{2g}\)
Correct answer is C
No explanation has been provided for this answer.