Q124 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Physics
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
- A10 m
- B14 m
- C18 m
- Correct answerD16 m
Solution
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).
$$ \begin{array}{rlrl} \text { When } & y & =3, x=4, \text { so } 3=4-\frac{g\left(4^2\right)}{u^2} \\ \Rightarrow & & \frac{16 g}{u^2} & =1 \\ \Rightarrow & & u^2 & =16 g \end{array} $$ 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).
Hence, the correct answer is option: 'D'.