Question

A man is standing at the top of 105 meter high tower. He throws a ball vertically upwards, with a velocity of 20m/s. After how many seconds will the ball pass him going downward? How long after its release will the ball reach the ground?

Answer 1

google pe kar sakta hai

Answer 2

u=20m/s
v=om/s
let g=10m/s2 
v=u+at
o=20+(-10)*t(a=g)
0  =  20-10t
10t=20
t=20/10=2sec
now, v^{2}=u^{2}+2-gh
          o    =20*20+2*-10*h
              0    =400-20h
             20h=400
             h=400/20=20m
hence total height ground to max height=105+20=125m