Final answer:
The initial velocity of the ball relative to the quarterback is approximately 19.72 m/s.
Step-by-step explanation:
The initial velocity of the ball relative to the quarterback can be found by adding the quarterback's velocity to the velocity of the ball relative to the ground.
Given that the quarterback is moving straight backward at a speed of 2.00 m/s, we can represent this velocity as -2.00 m/s since it is in the opposite direction of the positive x-axis.
The horizontal component of the ball's velocity can be found using the formula v_x = v * cos(theta), where v is the magnitude of the initial velocity and theta is the angle of the throw.
Similarly, the vertical component of the ball's velocity can be found using the formula v_y = v * sin(theta).
Given that the ball is thrown at an angle of 25.0° and caught at the same height, we can assume that the vertical component of the ball's velocity is 0 m/s.
Since the horizontal displacement is 18.0 m, we can use the equation x = v_x * t to find the time of flight of the ball. Rearranging the equation, we have t = x / v_x.
Plugging in the values, we have t = 18.0 m / (v * cos(25.0°)).
Solving for v, we have v = 18.0 m / (t * cos(25.0°)).
Now, substituting the value of t into the equation, we get v = 18.0 m / ((18.0 m / (v * cos(25.0°))) * cos(25.0°)).
Simplifying the equation, we have v = 18.0 m / (18.0 m / (v * cos(25.0°))).
Multiplying both sides of the equation by (v * cos(25.0°)), we get v * (v * cos(25.0°)) = 18.0 m.
Expanding the equation, we have v^2 * cos(25.0°) = 18.0 m.
Dividing both sides of the equation by cos(25.0°), we get v^2 = 18.0 m / cos(25.0°).
Taking the square root of both sides of the equation, we get v = sqrt(18.0 m / cos(25.0°)).
So, the initial velocity of the ball relative to the quarterback is approximately 19.72 m/s.