Physics forum is not helping?
So here is my ? again
I dont really want the answer im trying to understand
A hotair balloonist, rising vertically with a constant speed of 5.00m/s , releases a sandbag at the instant the balloon is 40.0m above the ground. After it is released, the sandbag encounters no appreciable air drag.
Compute the position (height above the ground) of the sandbag at 0.255s after its release.
please help just tried to solve this and cant get it right 
what does r stand for? sorry my teacher uses different variables. Here they are
X final
Xo
Vo
V final
The information that they gave you for the hot air balloonist rising at 5m/s is extra information and not necessary for your answer.
Your equations
r(t) = .5a(t)t^(2) + v(initial)t + r(initial)
v(t) = dr/dt = a(t)t + v(initial)
a(t) = dv/dt = a(t)
Given information:
r(initial) = 40m
v(initial) = 0 <--
0 because the initial speed of the sandbag when released is 0 and not 5m/s
t = 0.255
Can you pick which equations to use from here an plug in numbers?
Edit: r(t) is the same as x(t). They both mean position.
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