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

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$18$

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(C) The standard equation of trajectory for a projectile is given by $y = x 	an 	heta - \frac{g}{2u^2 \cos^2 	heta} x^2$. Comparing this with the given equation $y = 10x - \frac{5}{9}x^2$, we get: $	an 	heta = 10$ and $\frac{g}{2u^2 \cos^2 	heta} = \frac{5}{9}$. Given $g = 10 \ m/s^2$, we substitute it into the second equation: $\frac{10}{2u^2 \cos^2 	heta} = \frac{5}{9} \implies \frac{5}{u^2 \cos^2 	heta} = \frac{5}{9} \implies u^2 \cos^2 	heta = 9$. The range $R$ of a projectile is given by $R = \frac{2u^2 \sin 	heta \cos 	heta}{g}$. We can rewrite this as $R = \frac{2(u^2 \cos^2 	heta) 	an 	heta}{g}$. Substituting the values $u^2 \cos^2 	heta = 9$, $	an 	heta = 10$, and $g = 10 \ m/s^2$: $R = \frac{2 	imes 9 	imes 10}{10} = 18 \ m$.

Column $I$	Column $II$
$(A)$ Horizontal component of velocity	$(p)$ $5$ $SI$ unit
$(B)$ Vertical component of velocity	$(q)$ $10$ $SI$ unit
$(C)$ Horizontal displacement	$(r)$ $15$ $SI$ unit
$(D)$ Vertical displacement	$(s)$ $20$ $SI$ unit

The equation of trajectory of a projectile is $y = 10x - (5/9)x^2$. If we assume $g = 10 \ m/s^2$, the range of the projectile (in meters) is:

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