Question

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

• A. $2.8m$
• B. $10.3m$
• C. $5.1m$
• D. $2.5m$

Answer 1

The correct option is A $2.8m$

$\begin{array}{c}Accorrdingtoquestion.......................\\ Here,\\ Equationofpathofaprojectile-\\ V_x=10\cos 60=5m/s\\ V_y=10\cos 30=5\sqrt{3}m/s\\ Velocityaftert=1\sec \\ V_x=5m/s\\ \Rightarrow 1V_{y}1=(10-5\sqrt{3})m/s\\ Q_n=\frac{V^2}{R}\Rightarrow R=\frac{{V_x}^2+{V_y}^2}{an}|\begin{array}{c}\underset{–}{itisequationofparabola:}\\ y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta },\end{array}\\ \Rightarrow R=\frac{27+100+75-100\sqrt{3}}{10\cos \Theta }\\ \Rightarrow \Theta ={\tan }^{-1}\left(\frac{10-5\sqrt{3}}{5}\right)={15}^0\\ \therefore R=\frac{100(2-\sqrt{3})}{10\cos 15}=2.8m\\ SothecorrectoptionisA.\end{array}$