Using this formula:
u1(t+3) - 0.5*g*(t+3)^2 = u2*t - 0.5*g*t^2
From this I simplified as follows:
u1t +3u1-0.5*9.81*(t2+9) = u2*t - 0.5*9.81*t2
u1t +3u1-4.905*(t2+9) = u2*t - 4.905*t2
u1t +3u1-4.905*t2-4.905*9 = u2*t - 4.905*t2
The -4.905*t2 on both sides should cancel out...
If I use this as the time and recalculate the distance for each arrow it doesn't jive.
Through trial and error I found the meeting time to be 1.5345 seconds at a distance of 80.52 meters from the ground.
Not sure if I'm missing something or its just a minor error on my part.
So then:
u1t2+3u1+.5at22+4.5a=u2t2+.5at22
u1=40m/s
u2=60m/s
a=-9.81m/s2
The .5at22 on each side should cancel each other out leaving:
u1t2+3u1+4.5a=u2t2
Solving:
40t2+3(40)+4.5(-9.81)=60t2
40t2+120-44.145=60t2
20t2=75.855
t2=75.855/20
t2=3.79275 seconds
Homework Statement
Two arrows are fired straight up. The first is at 40 m/s and the second at 60 m/s and 3 seconds after arrow one. At what hieght will the arrows meet in the air.
Homework Equations
?
s=ut+1/2 at^2
The Attempt at a Solution
s1=s2
t2=t1+3...