A particle P is projected with velocity u_1 a | vedclass

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(B) For the particles to collide,they must reach the same point at the same time. Particle $P$ reaches its maximum height at the horizontal distance e

Vedclass · Accepted Answer

$u_1 = 2u_2$

Vedclass · Answer

(B) For the particles to collide,they must reach the same point at the same time. Particle $P$ reaches its maximum height at the horizontal distance equal to half of its range. At the maximum height,the vertical component of the velocity of particle $P$ becomes zero. However,for the particles to collide at the peak,the vertical velocity of particle $Q$ must match the vertical velocity of particle $P$ at the point of collision. Since particle $Q$ is projected from the point directly below the maximum height of $P$,it will reach the maximum height of $P$ at the same time $P$ reaches it if their vertical velocities are equal at that point. The vertical component of the initial velocity of $P$ is $u_{1y} = u_1 \sin 30^\circ$. For collision at the maximum height,the vertical velocity of $Q$ must be equal to the vertical component of $P$'s velocity at the start,or more precisely,the time taken to reach the peak must be the same. Since $v_y = u_y - gt$,at the peak $v_y = 0$,so $t = \frac{u_y}{g}$. For $P$,$t_P = \frac{u_1 \sin 30^\circ}{g}$. For $Q$,$t_Q = \frac{u_2}{g}$. Equating $t_P = t_Q$,we get $u_1 \sin 30^\circ = u_2$. $u_1 	imes \frac{1}{2} = u_2$. Therefore,$u_1 = 2u_2$.

$P$. Time of flight	$1$. $3$
$Q$. Time when particle is moving at an angle of $53^{\circ}$ with horizontal in upwards direction.	$2$. $10$
$R$. Range of the particle	$3$. $50$
$S$. Maximum height reached by the particle	$4$. $60$

$A$ particle $P$ is projected with velocity $u_1$ at an angle of $30^\circ$ with the horizontal. Another particle $Q$ is projected vertically upwards with velocity $u_2$ from the point directly below the maximum height of $P$. The condition for the collision of both particles is:

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