Answer:
The answer would be [tex]V_{0}=40,429 ft/s[/tex]
Explanation:
In this problem, we will use the range that is given and the equation of the flying time.
The ball leaves the bat at an angle so the speed of the ball has x and y vectors. So as i draw in the picture the speed has [tex]V_{x} =Vcos55[/tex] and [tex]V_{y} =Vsin55[/tex]
Also [tex]V_{x}[/tex] speed is constant through flying time
So to find the range we use the equation:
[tex]X_{range} =V_{x} t=Vcos55t[/tex]
Now, we know the range but to find V, we need the time t.
Total flying time can be found using the equation:
[tex]t_{flying} =\frac{2V_{y} }{g}[/tex]
If we put t in the first equation:
[tex]48=Vcos55\frac{2Vsin55}{g}[/tex]
Now since we also know g if we solve the equation we find [tex]V=40,429ft/s[/tex]
