Answer
a) |v|=8.18m/s, going up
bi) Height= 0.411m
bii) Height = 1152m
c) v_{max}=9.8m/s=g
d)v_{min}=g/2
Work Step by Step
Throughout, we will take downwards as positive
a) Of ball 2: 20=0 + 1/2 (g) (t-1)^{2}
t=3.0203s
Of ball 1: 20=u(3.0203) +1/2(g)(3.0203)^{2}
u=-8.178=-8.18m/s(3s.f). Negative direction indicates upwards initial velocity
bi) Of ball 1: -h=u(t) -1/2 gt^{2}
Ball 2: -h=1/2 g(t-1)^{2}
Given u=6 (upwards),
therefore,
-6(t)-1/2 gt^{2}=1/2 g(t-1)^{2}
t=1.289s
h=-6(1.289)+1/2(9.8)(0.289)^{2}=0.411m
bii) Similarly, v=0.3, we get t=15.3m
h=1152m
c)1/2g(t-1)^{2}=u(t)+1/2(g)(t)^{2}
we get, 0=ut+u+gt+g/2
differentiate w.r.t to t and you get u+g=0
u=-g
therefore Vmax=g
d) To not hit the ground at same time, they should meet again at top of building.
We have 2 options at t=0 and another at t=1 such that s=h
Equate them and we have: h= u(1)+1/2(g)(1)^{2}+h =u+g/2 = 0
Vmin=g/2
