Question

I shot two arrows into the air the first arrow had a speed of 10 m/sec , 1 second later I shot the 2nd arrow into the air with a speed of 15 m/sec.

Will they meet? If so, where?

Answer 1

First arrow:
V = Vo + g*Tr = 0
Tr = -Vo/g = -10/-10 = 1.0 s.=Rise time
So #2 is shot at the same time #1 starts falling.

h = Vo*Tr - 0.5g*Tr^2
h = 10*1 - 5*1^2 = 5 m.

ho-0.5g*t^2 = Vo*t + 0.5g*t^2
5 - 5*t^2 = 15*t - 5*t^2
5 - 5t^2 + 5t^2 = 15t
The 5t^2 terms cancel:
15t = 5
t = 0.333 s.

h = 5 - 5*0.333^2 = 4.45 m.
So they meet 4.45 m above the launching
point.