A particle `A` is projected from the ground with an initial velocity of `10m//s` at an angle of `60^(@)` with horizontal. From what height should an 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=10 m//s^(2)`.

Explanation

Let `u_(A) and u_(B)` are the initial speed of projection for A and B respectively
Horizontal component of velocity of A is 10 cos `60^(@)` or 5 m/s which is equal to the velocity of B in horizontal direction. They will collide at C, if the times of flight of particles are equal i.e. `t_(A) = t_(B)`
Since `t_(A) =(2u sin 60^(@))/(g) and h =(1)/(2) g_(B)^(2), and h =(1)/(2) gt_(B)^(2)," therefore "(2u_(A) sin 0)/(g) =sqrt((2h)/(g))`
or `h=(2u_(A)^(2) sin^(2)0)/(g) =(2(10)^(2) ((sqrt(3))/(2))^(2))/(10) =15 m`