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(D) The standard 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

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$5 \ m/s$

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(D) The standard 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 $y = x - \frac{2x^2}{5}$,we get:$	an 	heta = 1 \implies 	heta = 45^{\circ}$.Also,$\frac{g}{2u^2 \cos^2 	heta} = \frac{2}{5}$.Substituting $g = 10 \ m/s^2$ and $\cos^2 45^{\circ} = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$:$\frac{10}{2u^2 (1/2)} = \frac{2}{5}$.$\frac{10}{u^2} = \frac{2}{5}$.$2u^2 = 50 \implies u^2 = 25$.$u = 5 \ m/s$.

Column-$I$	Column-$II$
$(1)$ Vertical component of velocity is zero	$(a)$ Tangent to the parabolic path
$(2)$ Linear velocity	$(b)$ Maximum height point of the projectile trajectory

The trajectory of a projectile projected from the origin is given by the equation $y = x - \frac{2x^2}{5}$. The initial velocity of the projectile is:

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Column-$I$ Column-$II$

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