Answer:
[tex]\alpha =54.7[/tex]º
Explanation:
From the exercise we have our initial information
[tex]y=3.4m\\v_{o}=10m/s\\g=-9.8m/s^2[/tex]
When the balloon gets to the ceiling its velocity at that moment is 0 m/s. Being said that we can calculate velocity at the vertical direction
[tex]v_{y}^2=v_{oy}^2+ag(y-y_{o})[/tex]
Since [tex]v_{y}=0[/tex] and [tex]y_{o}=0[/tex]
[tex]0=v_{oy}^2-2(9.8m/s^2)(3.4m)[/tex]
[tex]v_{oy}=\sqrt{2(9.8m/s^2)(3.4m)}=8.16m/s[/tex]
Knowing that
[tex]v_{oy}=v_{o}sin\alpha[/tex]
[tex]sin\alpha =\frac{v_{oy} }{v_{o} }[/tex]
[tex]\alpha =sin^{-1}(\frac{v_{oy}}{v_{o}})=sin^{-1}(\frac{8.16m/s}{10m/s})=54.7[/tex]º
