Given Data:

• Initial velocity $v = 708 \, \text{m/s}$
• Launch angle $\theta = 36^\circ$
• Acceleration due to gravity $g = -10 \, \text{m/s}^2$
• Time at position 6: $t_6 = 52.00 \, \text{s}$
• From the graph, the range $x$ increases horizontally, and altitude $y$ changes vertically.

Finding the answers:

1. Downrange distance $x_6$:

   • The horizontal velocity component $v_x$ is constant throughout the motion: $v_x = v \cos(\theta)$ Substituting values: $v_x = 708 \cos(36^\circ)$ Using $\cos(36^\circ) \approx 0.809$: $v_x = 708 \times 0.809 \approx 573 \, \text{m/s}$

   • The horizontal distance $x$ traveled at time $t_6$ is: $x_6 = v_x \cdot t_6$ Substituting $v_x = 573 \, \text{m/s}$ and $t_6 = 52.00 \, \text{s}$: $x_6 = 573 \times 52.00 = 29,796 \, \text{m}$ Rounded to the nearest whole number: $x_6 = 29,800 \, \text{m}$

2. Altitude $y_6$ at position 6:

   • The vertical position $y$ is determined using the kinematic equation: $y = v_y \cdot t + \frac{1}{2} g t^2$ where $v_y = v \sin(\theta)$ is the initial vertical velocity.
     Substituting $v = 708 \, \text{m/s}$ and $\sin(36^\circ) \approx 0.588$: $v_y = 708 \sin(36^\circ) \approx 708 \times 0.588 \approx 416 \, \text{m/s}$

   • Now substituting into the vertical position formula: $y_6 = v_y t_6 + \frac{1}{2} g t_6^2$ Substituting values $v_y = 416 \, \text{m/s}$, $g = -10 \, \text{m/s}^2$, and $t_6 = 52.00 \, \text{s}$: $y_6 = 416 (52.00) + \frac{1}{2} (-10) (52.00)^2$

     First, calculate each term