A particle $A$ is projected from the ground with an initial velocity of $10m/s$ at an angle of ${60}^{\circ }$with horizontal. From what height $h$ should another particle $B$ be projected horizontally with velocity $5m/s$ so that both the particles collide in ground at point $C$ if both are projected simultaneously ($g=10m{s}^{-2}$)

The correct option is C $15m$
Find time of flight for both the particles $A$ and $B$.
Formula Used: $T=\frac{2u\sin \theta }{g}$
Given,
Velocity of $1^{st}$ projectile, $u=10m/s$
Angle of projection, $\theta ={60}^{\circ }$
Horizontal component of velocity of $A$ is $10\cos {60}^{\circ }$ or $5m/s$ which is equal to the velocity of $B$ in horizontal direction. They will collide at $C$ if time of flight of the particles are equal

Time of flight,
$T=\frac{2u\sin {60}^{\circ }}{g}=\frac{2\times 10\sin {60}^{\circ }}{10}=\sqrt{3}\text{sec}$
Find the height for $B$ from where it is being projected.
Formula Used: $s=ut+\frac{1}{2}g{t}^2$
Now velocity of $2^{nd}$ projectile in vertical direction, $u_y=0$
Time, $t=\sqrt{3}\text{sec}$
Height of fall, $s=u_{y}t+\frac{1}{2}g{t}^2$
Substituting the values, we get
$=0+\frac{1}{2}\times 10\times (\sqrt{3}{)}^2=15m$