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 meters per
second. What is the magnitude of the ball’s initial
velocity?
15.7 m/s
To find the magnitude of the ball's initial velocity, you can use the concept of vector components. The velocity of the ball can be broken down into its horizontal and vertical components.
Given:
Angle of the ball's initial velocity = 40.0°
Horizontal component of initial velocity = 12.0 m/s (velocity in the x-direction)
The horizontal component of velocity can be determined using the trigonometric function cosine.
Calculating the horizontal component:
Horizontal velocity (Vx) = Initial velocity * cos(angle)
Vx = 12.0 m/s * cos(40.0°)
Vx = 12.0 m/s * 0.766
Next, to find the magnitude of the ball's initial velocity, you can use the Pythagorean theorem, which states that the square of the hypotenuse (the magnitude of the velocity) is equal to the sum of the squares of the two sides (horizontal and vertical components).
Calculating the magnitude:
Magnitude of initial velocity (V) = sqrt(Vx^2 + Vy^2)
However, we still need to find the vertical component of the velocity (Vy).
The vertical component of velocity can be determined using the trigonometric function sine.
Calculating the vertical component:
Vertical velocity (Vy) = Initial velocity * sin(angle)
Vy = 12.0 m/s * sin(40.0°)
Vy = 12.0 m/s * 0.643
Now that we have both the horizontal and vertical components, we can find the magnitude of the ball's initial velocity.
Calculating the magnitude:
V = sqrt(Vx^2 + Vy^2)
V = sqrt((12.0 m/s * 0.766)^2 + (12.0 m/s * 0.643)^2)
Simplifying the calculation:
V = sqrt(108.144 + 94.272)
V = sqrt(202.416)
V ≈ 14.22 m/s
Therefore, the magnitude of the ball's initial velocity is approximately 14.22 meters per second.