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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Aicharya · Accepted Answer

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)

A body projected with some velocity at an angle $45^{\circ}$ with the horizontal from the origin in $X…

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