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(D) The equation of the trajectory of a projectile is given by:$y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$Given $	heta = 45^{\circ}$,$	a

Vedclass · Accepted Answer

$16$

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(D) The equation of the trajectory of a projectile is given by:$y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$Given $	heta = 45^{\circ}$,$	an 45^{\circ} = 1$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,so $\cos^2 45^{\circ} = \frac{1}{2}$.The equation becomes:$y = x - \frac{g x^2}{2 u^2 (1/2)} = x - \frac{g x^2}{u^2} \quad ... (i)$Since the projectile passes through the point $(4, 3) \ m$,we substitute $x = 4$ and $y = 3$ into equation $(i)$:$3 = 4 - \frac{g(4^2)}{u^2}$$3 = 4 - \frac{16g}{u^2}$$\frac{16g}{u^2} = 1 \Rightarrow u^2 = 16g$Substituting $u^2 = 16g$ back into equation $(i)$:$y = x - \frac{g x^2}{16g} = x - \frac{x^2}{16}$The horizontal range $R$ is the value of $x$ when $y = 0$ (at the landing point):$0 = x - \frac{x^2}{16}$$x(1 - \frac{x}{16}) = 0$Since $x 
eq 0$ at the landing point,$1 - \frac{x}{16} = 0$,which gives $x = 16 \ m$.Thus,the horizontal range is $16 \ m$.

$A$ body projected with some velocity at an angle $45^{\circ}$ with the horizontal from the origin in $XY$-plane passes through a point at $(4, 3) \ m$. Its horizontal range is (in $m$)

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