A particle P is projected with velocity u1 at an angle of 30° with the horizontal. Another particle Q is thrown vertically upwards with velocity u2 from a point vertically below the highest point of path of P. The necessary condition for the two particles to collide at the highest point is

The correct option is u1 = 2u2.

Let us analyze the motion of both particles to find the condition for their collision at the highest point of the path of particle P.

1. Motion of Particle P:
Particle P is projected with an initial velocity $u_1$ at an angle $\theta =30°$ with the horizontal.
The vertical component of the initial velocity of P is:
$u_{1y}=u_1\sin (30°)$
Since $\sin (30°)=\frac{1}{2}$, we have:
$u_{1y}=\frac{u_1}{2}$

At the highest point of its trajectory, the vertical velocity component of P becomes zero. The time $t$ taken by P to reach its highest point is given by:
$t=\frac{u_{1y}}{g}=\frac{u_1}{2g}$
The maximum height $H$ reached by particle P is:
$H=\frac{u_{1y}^2}{2g}=\frac{u_1^2}{8g}$

2. Motion of Particle Q:
Particle Q is thrown vertically upwards with an initial velocity $u_2$ from a point vertically below the highest point of the path of P. For the two particles to collide at this highest point, Q must reach the height $H$ at the exact same time $t$ that P reaches it.
Using the second equation of motion for Q:
$H=u_{2}t-\frac{1}{2}g{t}^2$

Substitute the expressions for $t$ and $H$ obtained from the motion of P into the equation for Q:
$\frac{u_1^2}{8g}=u_2\left(\frac{u_1}{2g}\right)-\frac{1}{2}g{\left(\frac{u_1}{2g}\right)}^2$

Simplify the term containing $t^2$:
$\frac{u_1^2}{8g}=\frac{u_1{u}_2}{2g}-\frac{u_1^2}{8g}$

Add $\frac{u_1^2}{8g}$ to both sides of the equation:
$\frac{u_1^2}{4g}=\frac{u_1{u}_2}{2g}$

Since $u_1\ne 0$ and $g\ne 0$, we can divide both sides by $\frac{u_1}{4g}$:
$u_1=2u_2$

Thus, the necessary condition for the collision to occur at the highest point is $u_1=2u_2$.