Answer:
The camera lands in t=2.91s with a velocity:
[tex]v=-30.45m/s[/tex]
Explanation:
The initial velocity of the camera is the same as the hot-air ballon:
[tex]v_{o}=-1.9m/s[/tex]
[tex]y_{o}=47m[/tex]
Kinematics equation:
[tex]v(t)=v_{o}-g*t[/tex]
[tex]y(t)=y_{o}+v_{o}t-1/2*g*t^{2}[/tex]
when the camera lands, y=0:
[tex]0=47-1.9t-4.91*t^{2}[/tex]
We solve this equation to find t:
t1=-3.29s This solution have not sense in our physical point of view
t2=2.91s
So, the camera lands in t=2.91s
We replace this value in v(t):
[tex]v=v_{o}-g*t=-1.9-9.81*2.91=-30.45m/s[/tex]
