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(B) The general equation of a projectile is $y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.Comparing this with the given equation $y = 12x -

Vedclass · Accepted Answer

$21.6$

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(B) The general equation of a projectile is $y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.Comparing this with the given equation $y = 12x - \frac{5}{9}x^2$:$1) 	an 	heta = 12$$2) \frac{g}{2 u^2 \cos^2 	heta} = \frac{5}{9}$Given the horizontal component of velocity $v_x = u \cos 	heta = 3 \ ms^{-1}$.Substitute $u \cos 	heta = 3$ into the second equation:$\frac{10}{2 (3)^2} = \frac{5}{9} \implies \frac{10}{18} = \frac{5}{9}$,which is consistent.To find the range $R$,we set $y = 0$ in the equation of motion:$0 = 12x - \frac{5}{9}x^2$$x(12 - \frac{5}{9}x) = 0$Since $x 
eq 0$,$12 = \frac{5}{9}x$$x = \frac{12 	imes 9}{5} = \frac{108}{5} = 21.6 \ m$.Thus,the range of the projectile is $21.6 \ m$.

The equation of motion of a projectile is $y = 12x - \frac{5}{9}x^2$. The horizontal component of velocity is $3 \ ms^{-1}$. Given that $g = 10 \ ms^{-2}$,the range of the projectile is .......... $m$.

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