Answer: 211.059 m
Explanation:
We have the following data:
[tex]\theta=70\°[/tex] The angle at which the ball leaves the bat
[tex]V_{o}=55 m/s[/tex] The initial velocity of the ball
[tex]g=-9.8 m/s^{2}[/tex] The acceleration due gravity
We need to find how far (horizontally) the ball travels in the air: [tex]x[/tex]
Firstly we need to know this velocity has two components:
Horizontally:
[tex]V_{ox}=V_{o}cos \theta[/tex] (1)
[tex]V_{ox}=55 m/s cos(70\°)=18.811 m/s[/tex] (2)
Vertically:
[tex]V_{oy}=V_{o}sin \theta[/tex] (3)
[tex]V_{oy}=55 m/s sin(70\°)=51.683 m/s[/tex] (4)
On the other hand, when we talk about parabolic movement (as in this situation) the ball 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 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] (5)
Isolating [tex]t[/tex]:
[tex]t=\frac{-V_{o}}{g}[/tex] (6)
[tex]t=\frac{-55 m/s}{-9.8 m/s^{2}}[/tex] (7)
[tex]t=5.61 s[/tex] (8)
Now that we have the time it takes to the ball to travel half of is path, we can find the total time [tex]T[/tex] it takes the complete parabolic path, which is twice [tex]t[/tex]:
[tex]T=2t=2(5.61 s)=11.22 s[/tex] (9)
With this result in mind, we can finally calculate how far the ball travels in the air:
[tex]x=V_{ox}T[/tex] (10)
Substituting (2) and (9) in (10):
[tex]x=(18.811 m/s)(11.22 s)[/tex] (11)
Finally:
[tex]x=211.059 m[/tex]
