Answer:
The magnitude of the ball's initial velocity is approximately 13.468 meters per second.
Explanation:
From Trigonometry we get that magnitude of the velocity is obtained from the following trigonometric relation:
[tex]\cos \theta = \frac{v_{x}}{v}[/tex] (Eq. 1)
Where:
[tex]v_{x}[/tex] - Magnitude of the horizontal component of the baseball, measured in meters per second.
[tex]v[/tex] - Magnitude of the velocity of the baseball, measured in meters per second.
[tex]\theta[/tex] - Angle of baseball above the horizontal, measured in sexagesimal degrees.
Then, we clear the magnitude of the velocity of the baseball:
[tex]v = \frac{v_{x}}{\cos \theta}[/tex]
If we know that [tex]v_{x} = 12\,\frac{m}{s}[/tex] and [tex]\theta = 30^{\circ}[/tex], then the magnitude of the velocity of the baseball is:
[tex]v = \frac{12\,\frac{m}{s} }{\cos 30^{\circ}}[/tex]
[tex]v \approx 13.468\,\frac{m}{s}[/tex]
The magnitude of the ball's initial velocity is approximately 13.468 meters per second.
