A particle projected vertically upward from a point A on the ground. It takes time t1 to reach a point B, but it still continues to move up. If it takes time t2 to reach the ground from point B. Then height of point B from the ground is ?

LET THE HIGHEST POINT WHERE THE BALL REACHES AFTER POINT B be C.TIME TAKEN BY BALL TO REACH B IS TO AND TIME TAKEN TO (REACH C FROM B+C to A) is =toWe know that Time taken to reach max.height is = time taken to reach ground from that point.Therefore ,Total time of flight=(t1+t2)Time taken to reach C from A=(t1+t2)÷2From A to C-v=u+(-g)t0=u-g (t1+t2)÷2u=(gt1+gt2)÷2using s=ut+0.5at^2 till bNow,by putting value of u,a=-g,t=t1we get-AB=(0.5)(g)(t1)(t2)