Question

A baseball is the road and angle of 30° above the horizontal. The horizontal component of the baseballs initial velocity is 12 m/s. What is the magnitude of the balls initial velocity

Answer 1

The magnitude of the ball's initial velocity is approximately 13.468 meters per second.

Step-by-step explanation:

From Trigonometry we get that magnitude of the velocity is obtained from the following trigonometric relation:

`\cos \theta = (v_(x))/(v)` (Eq. 1)

Where:

`v_(x)` - Magnitude of the horizontal component of the baseball, measured in meters per second.

`v` - Magnitude of the velocity of the baseball, measured in meters per second.

`\theta` - Angle of baseball above the horizontal, measured in sexagesimal degrees.

Then, we clear the magnitude of the velocity of the baseball:

`v = (v_(x))/(\cos \theta)`

If we know that
`v_(x) = 12\,(m)/(s)` and
`\theta = 30^(\circ)`, then the magnitude of the velocity of the baseball is:

`v = (12\,(m)/(s) )/(\cos 30^(\circ))`

`v \approx 13.468\,(m)/(s)`

The magnitude of the ball's initial velocity is approximately 13.468 meters per second.