Physics
posted by Boateng Emmanuel on .
Two particles Q and P are shot up vertically.P was first shot with velocity of40m/s and after4s.Q was also shot up.If upon meeting the velocity ofP is 15m/s.determine(1)where they met.(2)the velocity with which Q was shot up.Take g= 10m/s2

PARTICLE P:
Tr = (VfVo)/g,
Tr = (040) / 10 = 4.0 s. = Rise time.
hmax = Vo*t + 0.5g*t^2,
hmax = 40*4  5*4^2 = 80 m.
P is falling when Q is shot up:
1. h = hmax  (Vf^2Vo^2)/2g,
h = 80  ((15)^20) / 20 = 68.75 m.
Tf = (VfVo)/g = (150) / 10 = 1.5 s. =
Fall time. = Rise time(Tr) for Q.
PARTICLE Q:
2. h = Vo*t + 0.5g*t^2 = 8068.75 =11.25
Vo*1.5  5*(1.5)^2 = 11.25,
Vo*1.5  11.25 = 11.25,
Vo*1.5 = 22.5,
Vo = 22.5 / 1.5 = 15 m/s.