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A body projected with some velocity at an angle $45^{\circ}$ with the horizontal from the origin in $X Y$-plane passes through a point at $(4,3) \mathrm{m}$. Its horizontal range is
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The correct answer is:
16 m
Equation of trajectory of projectile is
$$
y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}
$$
Here, $\theta=45^{\circ}$
$$
\begin{array}{rlrl}
\text { When } & y & =3, x=4, \text { so } 3=4-\frac{g\left(4^2\right)}{u^2} \\
\Rightarrow & & \frac{16 g}{u^2} & =1 \\
\Rightarrow & & u^2 & =16 g
\end{array}
$$
Substituting the value of $u^2$ in Eq (i), we get
$$
\begin{aligned}
& y=x-\frac{g x^2}{16 g} \\
& y=x-\frac{x^2}{16}
\end{aligned}
$$
How at maximum $x, y=0$.
Hence, $\quad x-\frac{x^2}{16}=0 \Rightarrow x\left(1-\frac{x}{16}\right)=0$
$\Rightarrow x=0$ (initial position) and $x=16 \mathrm{~m}$ (final position).
$$
y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}
$$
Here, $\theta=45^{\circ}$
$$
\begin{array}{rlrl}
\text { When } & y & =3, x=4, \text { so } 3=4-\frac{g\left(4^2\right)}{u^2} \\
\Rightarrow & & \frac{16 g}{u^2} & =1 \\
\Rightarrow & & u^2 & =16 g
\end{array}
$$
Substituting the value of $u^2$ in Eq (i), we get
$$
\begin{aligned}
& y=x-\frac{g x^2}{16 g} \\
& y=x-\frac{x^2}{16}
\end{aligned}
$$
How at maximum $x, y=0$.
Hence, $\quad x-\frac{x^2}{16}=0 \Rightarrow x\left(1-\frac{x}{16}\right)=0$
$\Rightarrow x=0$ (initial position) and $x=16 \mathrm{~m}$ (final position).
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