A particle A is projected from the ground with an initial velocity of  10m/s at an angle of  600 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=10m/s2.

Both the particles will meet at C, if their time of flight is the same.  The time Of flight of A is, T=2u  sin  θg T=2×10×   sin  60010 T=3second For vertical downward motion of particle B from B to C, we have h=12gT2         substitute T=3second h=12×10×32 h=15m=height from where ball B should be horizontally projected