A particle is projected vertically upwards from a point A on the ground. It takes time t1 to reach a point B but it still continues to move up. It takes further time t2 to reach ground from point B. Then height of B from gound is what? (Ans: 1/2 g t1 t2)

LET THE HIGHEST POINT WHERE THE BALL REACHES AFTER POINT B be C. TIME TAKEN BY BALL TO REACH B IS t1 AND TIME TAKEN TO (REACH C FROM B+C to A) is =t2. We 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)÷2 From A to C -; v=u+(-g)t 0=u-g (t1+t2)÷2u u=(gt1+gt2)÷2 using, s=ut+0.5at^2 till bNow,by putting value of u,a=-g,t=t1we get-AB=(0.5)(g)(t1)(t2)

Dear Student,
Please find below the solution to your problem.

Time taken for the particle to reach highest point is t1+t22t1+t22
(v=u−gt)(v=u−gt)
Therefore initial velocity=u=g(t1+t22)=u=g(t1+t22)
Therefore heigh of B from ground is
h=ut+12h=ut+12at2at2
h=ut1−12h=ut1−12gt12gt12
=g(t1+t22)=g(t1+t22)t1−12t1−12gt21gt12
h=12h=12gt1t2

Thanks and Regards