Consider the motion of a soccer ball that is kicked from the point \( \left\langle x_{0}, y_{0}\right\rangle=\langle 0,0\rangle \) with an initial velocity of \( \left\langle u_{0}, v_{0}\right\rangle=\langle 30,8\rangle \mathrm{m} / \mathrm{s} \). Assume the \( x \)-axis is horizontal, the positive \( y \)-axis is vertical (opposite g), the ground is horizontal, and only the gravitational force acts on the object. a. Find the velocity and position vectors for \( t \geq 0 \). b. Graph the trajectory. c. Determine the time of flight and range of the soccer ball. d. Determine the maximum height of the soccer ball. a. The velocity vector is \( \mathbf{v}(\mathrm{t})= \)

To analyze the motion of the soccer ball, we can decompose the problem into its horizontal and vertical components considering the initial velocity and the effects of gravity. **Velocity Vector:** The horizontal component of the velocity remains constant, while the vertical component changes due to gravity. The velocity vector \( \mathbf{v}(t) \) can be expressed as: \[ \mathbf{v}(t) = \langle u_0, v_0 - gt \rangle = \langle 30, 8 - 9.81t \rangle \text{ m/s} \] where \( g \approx 9.81 \text{ m/s}^2 \) is the acceleration due to gravity. **Position Vector:** To find the position vector \( \mathbf{r}(t) \), we integrate the velocity components: \[ \mathbf{r}(t) = \langle x_0 + u_0 t, y_0 + v_0 t - \frac{1}{2}gt^2 \rangle = \langle 0 + 30t, 0 + 8t - \frac{1}{2}(9.81)t^2 \rangle = \langle 30t, 8t - 4.905t^2 \rangle \] **Trajectory Graph:** To graph the trajectory, you can visualize the path of the soccer ball by plotting the position vector over time. The horizontal distance (x) increases linearly, while the vertical distance (y) follows a parabolic path due to the influence of gravity. **Time of Flight and Range:** To determine the time of flight, set \( y = 0 \): \[ 8t - 4.905t^2 = 0 \implies t(8 - 4.905t) = 0 \implies t = 0 \text{ (initial time)} \text{ or } t = \frac{8}{4.905} \approx 1.63 \text{ seconds} \] The range \( R \) can be calculated with the time of flight: \[ R = u_0 \cdot t = 30 \cdot 1.63 \approx 48.9 \text{ meters} \] **Maximum Height:** To find the maximum height \( H \), first, calculate the time at which the vertical velocity becomes zero: \[ 0 = v_0 - gt \implies t = \frac{v_0}{g} = \frac{8}{9.81} \approx 0.816 \text{ seconds} \] Substituting this back into the y-component of the position: \[ H = 8(0.816) - 4.905(0.816^2) \approx 3.27 \text{ meters} \] Thus, the elements of the ball's motion are captured in these calculations, making it an excellent illustration of projectile motion in physics!