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

Answer 1

PARTICLE P:

Tr = (Vf-Vo)/g,
Tr = (0-40) / -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^2-Vo^2)/2g,
h = 80 - ((15)^2-0) / 20 = 68.75 m.
Tf = (Vf-Vo)/g = (15-0) / 10 = 1.5 s. =
Fall time. = Rise time(Tr) for Q.

PARTICLE Q:

2. h = Vo*t + 0.5g*t^2 = 80-68.75 =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.

Answer 2

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