A body is projected at t = 0 with a velocity | vedclass

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Question

$\begin{array}{l}
at\,t = 1\
\,{u_x} = 5,\,{u_y} = 5\sqrt 3 \
{v_y} = 5\sqrt 3  - 10\,\,;\,\,\,{v_x} = 5\
	an \,	heta \, =  - 2\lef

Vedclass · Accepted Answer

$2.8$

Vedclass · Answer

$\begin{array}{l}
at\,t = 1\
\,{u_x} = 5,\,{u_y} = 5\sqrt 3 \
{v_y} = 5\sqrt 3  - 10\,\,;\,\,\,{v_x} = 5\
	an \,	heta \, =  - 2\left( {2 - \sqrt 3 } ight) \Rightarrow 	heta  =  - {30^ \circ }\
R = \frac{{{v^2}}}{{a \bot }} = \frac{{{{10}^2}}}{{\left( {10\,\cos \,{{30}^ \circ }} ight)}}
\end{array}$

$\begin{array}{l}
\,\, = \frac{{10}}{{\sqrt 3 }} 	imes 2\, = \frac{{20}}{{\sqrt 3 }}m\
\frac{{{5^2} + {{\left( {10 - 5\sqrt 3 } ight)}^2}}}{{10\,\cos 	heta }} = \frac{{200 - 100\sqrt 3 }}{{10 	imes 0.965}} = 2.8m
\end{array}$

A body is projected at $t = 0$ with a velocity $10\,ms^{-1}$ at an angle of $60^o$ 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 $g = 10\,ms^{-2},$ the radius of $R$ is ........ $m$

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