A baseball is thrown at an angle of 40.0 above the horizontal.The horizontal component of the baseball's initial velocity is 12.0 m/s^2 . What is the magnitude of the object's acceleration?
To find the magnitude of the object's acceleration, we first need to determine its vertical and horizontal components of velocity.
Given:
- Angle of the baseball's initial velocity: 40.0°
- Horizontal component of the baseball's initial velocity: 12.0 m/s
Using trigonometry, we can find the vertical component of velocity:
Vertical component of velocity (Vy) = Initial velocity (V) * sin(angle)
Vy = 12.0 m/s * sin(40.0°)
Vy ≈ 7.74 m/s
The vertical component of velocity represents the object's initial vertical velocity. Since gravity affects the vertical motion of the object, the acceleration due to gravity needs to be considered.
Acceleration due to gravity (g) is approximately 9.8 m/s², acting downwards.
Therefore, the magnitude of the object's acceleration can be calculated as:
Acceleration (a) = √(ax² + ay²)
Where:
ax = Acceleration in the horizontal direction (which is 0 since there is no horizontal acceleration)
ay = Acceleration in the vertical direction (which is -g, considering downward motion)
Plugging in the values:
a = √(0² + (-9.8 m/s²)²)
a ≈ 9.8 m/s²
Hence, the magnitude of the object's acceleration is approximately 9.8 m/s².