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

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(B) Initial velocity components: $u_x = 10 \cos 60^\circ = 5\,ms^{-1}$ and $u_y = 10 \sin 60^\circ = 5\sqrt{3}\,ms^{-1}$.At $t = 1\,s$,the velocity co

Vedclass · Accepted Answer

$2.8$

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(B) Initial velocity components: $u_x = 10 \cos 60^\circ = 5\,ms^{-1}$ and $u_y = 10 \sin 60^\circ = 5\sqrt{3}\,ms^{-1}$.At $t = 1\,s$,the velocity components are:$v_x = u_x = 5\,ms^{-1}$$v_y = u_y - gt = 5\sqrt{3} - 10(1) = 5\sqrt{3} - 10\,ms^{-1}$.The speed $v$ at $t = 1\,s$ is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{5^2 + (5\sqrt{3} - 10)^2} = \sqrt{25 + (75 + 100 - 100\sqrt{3})} = \sqrt{200 - 100\sqrt{3}} \approx \sqrt{200 - 173.2} = \sqrt{26.8} \approx 5.17\,ms^{-1}$.The radius of curvature is given by $R = \frac{v^2}{a_{\perp}}$,where $a_{\perp}$ is the component of acceleration perpendicular to the velocity.The angle $	heta$ that the velocity vector makes with the horizontal is $	an 	heta = \frac{v_y}{v_x} = \frac{5\sqrt{3} - 10}{5} = \sqrt{3} - 2 \approx -0.268$. Thus,$	heta \approx -15^\circ$.The acceleration $g$ acts vertically downwards. The component of $g$ perpendicular to the velocity is $a_{\perp} = g \cos 	heta$.$R = \frac{v^2}{g \cos 	heta} = \frac{200 - 100\sqrt{3}}{10 \cos(-15^\circ)} = \frac{26.8}{10 	imes 0.966} \approx 2.77\,m \approx 2.8\,m$.

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

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