A hot air balloon ascending at a rate of 10m/s drops a package when the balloon reaches a height of 73 m. Find time to ground. Find velocity when package hits ground.
The balloon and package, in this case, can be treated as a point particles and as such can be manipluated with genral kinematics equations.
I apologize for my inability to display mathematical symbols when creating equations....
X = X(initial) + V(initial)t + 1/2at^2
V(final)^2 = V(initial)^2 + at
The Attempt at a Solution
I simply used the quadratic formula with the first equation to solve for t because the equation is already in terms of t. This produced:
-10(+-)sq.rt(100+1460) / -10
I chose the negative value of b^2 - 4ac to produce:
t = -49.49/-10 = 4.949s
Secondly, I used this time in the V(final) = V(initial) + at equation:
V(final) = 10 + (-10*4.949) = -39.49m/s
Am I thinking about this scenario correctly enough so that I am using the appropriate equations?? The answers, I am told, are incorrect.