A body P is projected at an angle of 30^{circ | vedclass

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Question

(B) For body $P$, the angle of projection with the horizontal is $	heta_P = 30^{\circ}$.For body $Q$, the angle of projection with the vertical is $3

Vedclass · Accepted Answer

$1: 6$

Vedclass · Answer

(B) For body $P$, the angle of projection with the horizontal is $	heta_P = 30^{\circ}$.For body $Q$, the angle of projection with the vertical is $30^{\circ}$, so the angle with the horizontal is $	heta_Q = 90^{\circ} - 30^{\circ} = 60^{\circ}$.The horizontal range is given by $R = \frac{u^2 \sin 2	heta}{g}$.Given $\frac{R_P}{R_Q} = \frac{1}{2}$, we have $\frac{u_P^2 \sin(2 	imes 30^{\circ}) / g}{u_Q^2 \sin(2 	imes 60^{\circ}) / g} = \frac{1}{2}$.Since $\sin 60^{\circ} = \sin 120^{\circ} = \frac{\sqrt{3}}{2}$, the ratio becomes $\frac{u_P^2}{u_Q^2} = \frac{1}{2}$, so $u_P^2 : u_Q^2 = 1 : 2$.The maximum height is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.Therefore, $\frac{H_P}{H_Q} = \frac{u_P^2 \sin^2 30^{\circ}}{u_Q^2 \sin^2 60^{\circ}} = \left(\frac{u_P^2}{u_Q^2}ight) 	imes \left(\frac{\sin 30^{\circ}}{\sin 60^{\circ}}ight)^2$.Substituting the values: $\frac{H_P}{H_Q} = \left(\frac{1}{2}ight) 	imes \left(\frac{1/2}{\sqrt{3}/2}ight)^2 = \frac{1}{2} 	imes \left(\frac{1}{\sqrt{3}}ight)^2 = \frac{1}{2} 	imes \frac{1}{3} = \frac{1}{6}$.Thus, the ratio $H_P : H_Q = 1 : 6$.

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