A projectile A is thrown at an angle 30^{circ | vedclass

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(C) Let the projectile $A$ be thrown with velocity $v_1$ at an angle $	heta = 30^{\circ}$ with the horizontal.The time taken by projectile $A$ to rea

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) Let the projectile $A$ be thrown with velocity $v_1$ at an angle $	heta = 30^{\circ}$ with the horizontal.The time taken by projectile $A$ to reach its highest point is $t = \frac{v_1 \sin 	heta}{g}$.At the highest point,the vertical velocity of projectile $A$ becomes zero,and it only has a horizontal velocity component $v_x = v_1 \cos 	heta$.For the two projectiles to collide,they must reach the same point at the same time. Since $B$ is thrown from a point $Q$ vertically below the highest point of $A$,the horizontal distance between them is covered by $A$ in time $t$.For a collision to occur at the highest point of $A$,the vertical displacement of $B$ in time $t$ must equal the maximum height $H$ of projectile $A$.Maximum height of $A$ is $H = \frac{v_1^2 \sin^2 	heta}{2g}$.For projectile $B$,the displacement in time $t$ is $y = v_2 t - \frac{1}{2} g t^2$.Substituting $t = \frac{v_1 \sin 	heta}{g}$ into the equation for $y$:$y = v_2 \left( \frac{v_1 \sin 	heta}{g} ight) - \frac{1}{2} g \left( \frac{v_1 \sin 	heta}{g} ight)^2 = \frac{v_2 v_1 \sin 	heta}{g} - \frac{v_1^2 \sin^2 	heta}{2g}$.Setting $y = H$:$\frac{v_2 v_1 \sin 	heta}{g} - \frac{v_1^2 \sin^2 	heta}{2g} = \frac{v_1^2 \sin^2 	heta}{2g}$.$\frac{v_2 v_1 \sin 	heta}{g} = \frac{v_1^2 \sin^2 	heta}{g}$.$v_2 = v_1 \sin 	heta$.Given $	heta = 30^{\circ}$,$v_2 = v_1 \sin 30^{\circ} = v_1 \left( \frac{1}{2} ight)$.Therefore,$\frac{v_2}{v_1} = \frac{1}{2}$.

Column-$I$	Column-$II$
$(1)$ Vertical component of velocity is zero	$(a)$ Tangent to the parabolic path
$(2)$ Linear velocity	$(b)$ Maximum height point of the projectile trajectory

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