The equation of a projectile is y = sqrt{3}x | vedclass

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(B) The standard equation 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

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$2\sqrt{10} \, m/s$

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(B) The standard equation of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Comparing this with the given equation $y = \sqrt{3}x - \frac{x^2}{2}$,we get:$	an 	heta = \sqrt{3} \implies 	heta = 60^\circ$.Also,$\frac{g}{2u^2 \cos^2 	heta} = \frac{1}{2}$.Taking $g = 10 \, m/s^2$ and $	heta = 60^\circ$,we have $\cos 60^\circ = 1/2$,so $\cos^2 60^\circ = 1/4$.Substituting these values: $\frac{10}{2u^2(1/4)} = \frac{1}{2}$.$\frac{10}{u^2/2} = \frac{1}{2} \implies \frac{20}{u^2} = \frac{1}{2}$.$u^2 = 40 \implies u = \sqrt{40} = 2\sqrt{10} \, m/s$.

The equation of a projectile is $y = \sqrt{3}x - \frac{x^2}{2}$. The velocity of projection is:

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