A particle A is projected from the ground with an initial velocity of  10m/s at an angle of  ${60}^0$ 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  $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2.$

  $$\begin{gathered} Both the particles will meet at C, if their time of flight is the same. \\ The time Of flight of A is, \end{gathered}$$

$T=\frac{2u\sin \theta }{g}$

$T=\frac{2\times 10\times \sin {60}^0}{10}$
$$\begin{gathered} T=\sqrt{3}second \\ For vertical downward motion of particle B from B to C, we have \end{gathered}$$

$h=\frac{1}{2}g{T}^2 substitute T=\sqrt{3}second$

$$\begin{gathered} h=\frac{1}{2}\times 10\times {\left(\sqrt{3}\right)}^2 \\ h=15m=height from where ball B should be horizontally projected \end{gathered}$$