Answer: 7.734 m/s
Explanation:
We have the following data:
[tex]\theta=35\°[/tex] The angle at which the water ballon was thrown
[tex]x=10 m[/tex] The horizontal distance of the water ballon
[tex]g=-9.8 m/s^{2}[/tex] The acceleration due gravity
We need to find the initial velocity [tex]V_{o}[/tex] at which the water ballon was thrown, and we can find it by the following equation:
[tex]x=V_{o}cos \theta T[/tex] (1)
Where [tex]T=2t[/tex] is the total time the water ballon is on air
On the other hand, when we talk about parabolic motion (as in this situation) the water ballon reaches its maximum height just in the middle of this parabola, when [tex]V=0[/tex] and the time [tex]t[/tex] is half the time [tex]T[/tex] it takes the complete parabolic path.
So, if we use the following equation, we will find [tex]t[/tex]:
[tex]V=V_{o}+gt=0[/tex] (2)
Isolating [tex]t[/tex]:
[tex]t=\frac{-V_{o}}{g}[/tex] (3)
Remembering [tex]T=2t[/tex]:
[tex]T=2\frac{-V_{o}}{g}[/tex] (4)
Substituting (4) in (1):
[tex]x=V_{o}cos \theta (2\frac{-V_{o}}{g})[/tex] (5)
Isolating [tex]V_{o}[/tex]:
[tex]V_{o}=\sqrt{\frac{x g}{-2 cos \theta}}[/tex] (6)
[tex]V_{o}=\sqrt{\frac{(10 m)(9.8 m/s^{2})}{-2 cos(35\°)}}[/tex] (7)
Finally:
[tex]V_{o}=7.734 m/s[/tex]
