The vertical displacement (y in metre) of a p | vedclass

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(B) The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Comparing this with the given equ

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$\sqrt{3} \ s$

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(B) The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Comparing this with the given equation $y = \sqrt{3}x - 0.2x^2$:$1$. $	an 	heta = \sqrt{3} \implies 	heta = 60^\circ$.$2$. $\frac{g}{2u^2 \cos^2 	heta} = 0.2$.Given $g = 10 \ ms^{-2}$ and $\cos 60^\circ = 0.5$,we have $\frac{10}{2u^2 (0.5)^2} = 0.2$.$\frac{10}{2u^2 (0.25)} = 0.2 \implies \frac{10}{0.5u^2} = 0.2 \implies \frac{20}{u^2} = 0.2 \implies u^2 = 100 \implies u = 10 \ ms^{-1}$.The time of flight $T$ is given by $T = \frac{2u \sin 	heta}{g}$.$T = \frac{2 	imes 10 	imes \sin 60^\circ}{10} = 2 	imes \frac{\sqrt{3}}{2} = \sqrt{3} \ s$.

The vertical displacement ($y$ in metre) of a projectile in terms of its horizontal displacement ($x$ in metre) is given by $y = (\sqrt{3}x - 0.2x^2)$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ ms^{-2}$)

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