The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is 20ms^-1, its time of flight is (g = 10 ms^-2)

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$2 \sqrt{2} s$

1/2 s

$\frac{1}{\sqrt{2}} s$

answer is B.

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Detailed Solution

KE = PE
$\begin{array}{l} \frac{1}{2} {mu}^{2} \cos^{2} θ = \frac{1}{2} {mu}^{2} \sin^{2} θ \\ Tan θ = 1, θ = 45^{0} \\ T = \frac{2 usin θ}{g} = \frac{2 \times 20 \times \frac{1}{\sqrt{2}}}{10} = 2 \sqrt{2} \sec \end{array}$

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