From top of a tower a stone is thrown up and it reaches the ground in time t1. A second stone is thrown down with the same speed and it reaches the ground in t2. A third stone is released from rest and it reaches the ground in time t3 then t3= ?solution)
let the intial speed be ''u'' and height be ''h''.
h(-j)=u(j)*t1+0.5g(-j)*t1^{2}
Case(2)
h(-j)=u(-j)*t2+0.5g(-j)*t2^{2}
Case(3)
h(-j)=0.5g(-j)*t3^{2}
solve case 1 and case 2 for t1 and t2 (quadratic) .
its very easy quest. we just need to find the values in u,g,and v(final velocity,this is constant for the first two cases)
for first throw.
t1= 2u/g + v-u/g=v+u/g
for second throw.
t2=v-u/g
for third release.
t3=u/g
so t3=(t1-t2)/2g