Final answer:
The initial speed of the ball relative to the ground is approximately 15 m/s. It takes approximately 9 seconds for the ball to reach the receiver. The maximum height above the point of release is approximately 3.86 meters.
Step-by-step explanation:
To determine the initial speed of the ball relative to the ground, we need to analyze the horizontal and vertical components of its motion separately. The ball's initial horizontal speed will be the same as the quarterback's speed since it is moving straight backward. Therefore, its initial horizontal speed relative to the ground is 2.00 m/s.
Next, we can use the vertical component of the ball's motion to find its initial vertical speed. The ball is thrown at an angle of 25° relative to the ground, so its vertical speed can be calculated using the formula v = v₀sinθ, where v₀ is the initial speed and θ is the angle. Therefore, the ball's initial vertical speed relative to the ground is v₀sin25°.
Finally, we can find the ball's initial speed relative to the ground by combining its horizontal and vertical speeds using the Pythagorean theorem. The initial speed of the ball relative to the ground is given by the equation:
v = sqrt((2.00 m/s)^2 + (v₀sin25°)^2).
By calculating this equation, we can determine that the initial speed of the ball relative to the ground is approximately 15 m/s.
To find the time it takes for the ball to reach the receiver, we can use the horizontal distance between them and the initial horizontal speed of the ball. The time can be calculated using the equation:
t = d / v₀, where t is the time, d is the horizontal distance, and v₀ is the initial horizontal speed. Plugging in the values, we have:
t = 18.0 m / 2.00 m/s
By solving this equation, we find that it takes approximately 9 seconds for the ball to reach the receiver.
The maximum height of the ball above its point of release can be determined using the vertical motion component. Since the ball is caught at the same height as it is released, the maximum height is equal to the initial height. The formula for the maximum height of an object in projectile motion is:
h_max = (v₀*sinθ)^2 / (2*g), where h_max is the maximum height, v₀ is the initial vertical speed, θ is the angle, and g is the acceleration due to gravity. Plugging in the values, we have:
h_max = (v₀*sin25°)^2 / (2*9.8 m/s²)
By calculating this equation, we find that the maximum height of the ball above its point of release is approximately 3.86 meters.