a stone is projected upwards vertically with initial speed u m/s a second stone is thrown upward from the same point with the same initial speed T seconds later than the first one. prove that they collide at a height of (4u^2-g^2t^2)/8g meters above the point of projection. to work this out i used uvast equations. for the first particle u=u s=h a=-g t= k for the second particle u=u s=h a=-g and t= k-t . the second particle has been in the air for t seconds less than the first particle thats why its time= k-t . k stands for the amount of time the first particle hhas been in the air. anyway i worked these out using s=ut+.5at^2. and i dont know where to go from here.