A quarterback is trying to throw a football to a stationary receiver. The receiver is not directly in front of him, so the total velocity of the ball and the velocity of the quarterback has to be 20 m/s at an angle of 21.4 degrees from the vertical direction as shown in the sketch. If the quarterback is running at 8.9 m/s to his right (+x direction), how fast does the ball need to be thrown in m/s? In other words, what is the magnitude of the velocity of the ball?
To find the magnitude (speed) of the velocity at which the ball needs to be thrown, we can use vector addition. We know that the total velocity of the ball and the quarterback needs to be 20 m/s, and we also know the velocity of the quarterback is 8.9 m/s to the right.
First, let's break down the velocity of the ball into its horizontal and vertical components. We'll call the horizontal component Vx and the vertical component Vy.
The vertical component Vy can be found using the equation:
Vy = V * sin(θ)
where V is the magnitude of the velocity of the ball, and θ is the angle from the vertical direction (which is 21.4 degrees in this case).
The horizontal component Vx is equal to the velocity of the quarterback, which is 8.9 m/s to the right in the +x direction.
Now, let's use vector addition to find the magnitude of the velocity of the ball. The magnitude of the velocity is given by the equation:
V = sqrt(Vx^2 + Vy^2)
Substituting the known values, we have:
V = sqrt((8.9)^2 + (20 * sin(21.4))^2)
V = sqrt(79.21 + 78.76)
V = sqrt(157.97)
V ≈ 12.57 m/s
Therefore, the magnitude of the velocity at which the ball needs to be thrown is approximately 12.57 m/s.