A body is projected at $t=0$ with a velocity $10 \mathrm{~ms}^{-1}$ at an angle of $60^{\circ}$ with the horizontal. The radius of curvature of its trajectory at $t=1 s$ is $R$. Neglecting air resistance and taking acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$, the value of $R$ is :

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Aicharya · Accepted Answer

Step 1. Given Data: Initial velocity, $u=10 \mathrm{~m} / \mathrm{s}$ Angle of projection, $\theta=60^{\circ}$ Radius of curvature at $t=1 s$ is $R$ Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}$ Step 2. Finding the radius of curvature: Resolving the velocity into $x$ and $y$ components, at any time $t$ $ \begin{aligned} & v_x=u \cos 60^{\circ}=10 \times \frac{1}{2} \\ & v_x=5 \mathrm{~m} / \mathrm{s} \\ & v_y=u \sin 60^{\circ}+a t=\frac{10 \sqrt{3}}{2}-10 t \\ & v_y=|5 \sqrt{3}-10 t| \end{aligned} $

At $t=1 s$ $ v_y=|5 \sqrt{3}-10| $ Angle made by body with respect to horizontal axis at $t=1 \mathrm{~s}$ $ \begin{aligned} & \tan \alpha=\left|\frac{v_y}{v_x}\right|=\left|\frac{5 \sqrt{3}-10}{5}\right|, \\ & \tan \alpha==|\sqrt{3}-2| \\ & \alpha=\tan ^{-1}(|\sqrt{3}-2|)=15^{\circ} \end{aligned} $ The radius of curvature same as range at time $t$, which is given by $ \begin{aligned} R & =\frac{v^2}{g \cos \alpha}=\frac{v_x^2+v_y^2}{g \cos \alpha} \\ R & =\frac{5^2+(5 \sqrt{3}-10)^2}{10 \times \cos 15^o} \\ R & =\frac{25+75+100-100 \sqrt{3}}{10 \times \cos 15^{\circ}} \\ R & =\frac{26.79}{9.65}=2.77 \\ R & =2.77 m \approx 2.8 m \end{aligned} $ Hence, option $C$ is correct

A body is projected at $t=0$ with a velocity $10 \mathrm{~ms}^{-1}$ at an angle of $60^{\circ}$ with the…

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