An elevator departs the ground at t=0 With uniform velocity. At T1 a boy drops a marbel through the floor of the lift that falls under constant acceleration g=9.8 ms^-2. If it takes marbel T2 time to fall down, what was the height at the T1.
h= vT2 -.5g(T2)^2
The Attempt at a Solution
I simply replaced v in second equation by h/T1. And i got the h in terms of T1, T2. But the problem is, there is a clue given with this problem in the book. That is when T1=T2=4 h=39.2 now, in my equation, T1=T2 gives a height 0.