[tex]\begin{gathered}\begin{gathered}{\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} \\\end{gathered}\end{gathered}[/tex] To determine how far the cannonball will travel before hitting the ground, we can analyze the problem using the principles of projectile motion. Given Data Initial velocity ([tex]v_0[/tex]) = 50 m/s Angle of projection ([tex]\theta[/tex]) = 45° Acceleration due to gravity ([tex]g[/tex]) = 9.81 m/s² (downward) Resolve the Initial Velocity into Components The initial velocity can be broken down into horizontal and vertical components: Horizontal Component ([tex]v_{0x}[/tex]): [tex]v_{0x} = v_0 \cdot \cos(\theta) = 50 \cdot \cos(45^\circ) = 50 \cdot \frac{\sqrt{2}}{2} \approx 35.36 \text{ m/s}[/tex]Vertical Component ([tex]v_{0y}[/tex]): [tex]v_{0y} = v_0 \cdot \sin(\theta) = 50 \cdot \sin(45^\circ) = 50 \cdot \frac{\sqrt{2}}{2} \approx 35.36 \text{ m/s}[/tex] Calculate the Time of Flight To find the time the cannonball is in the air, we can use the vertical motion equations. The time to reach the highest point can be calculated using: [tex]t_{\text{up}} = \frac{v_{0y}}{g} = \frac{35.36}{9.81} \approx 3.6 \text{ s}[/tex] Since the projectile takes the same amount of time to go up as it does to come down, the total time of flight (T) is: [tex]T = 2 \cdot t_{\text{up}} \approx 2 \cdot 3.6 \approx 7.2 \text{ s}[/tex] Calculate the Horizontal Range The horizontal distance (range) traveled can be calculated using the horizontal velocity and the total time of flight: [tex]\text{Range} = v_{0x} \cdot T = 35.36 \text{ m/s} \cdot 7.2 \text{ s} \approx 254.59 \text{ m}[/tex] Conclusion The cannonball will travel approximately 254.59 meters before hitting the ground. [tex]\sf\color{green}{⊱⋅ ────────────────────── ⋅⊰}[/tex][tex]\begin{gathered} \boxed{\begin{array}{cc} \sf \footnotesize \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ···\: ʚ\: \: \: \: \: \: Hope\:it\:helps \: \: \: \: \: ɞ \:··· \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\footnotesize\:\# CarryOnLearning \\ \sf\footnotesize ૮₍´˶ \: • \: . \: • \: ⑅ ₎ა \: \leadsto \footnotesize\sf\color{purple} \underline{Study\:Well!}\end{array}}\end{gathered}[/tex][tex]\sf\color{green}{⊱⋅ ────────────────────── ⋅⊰}[/tex][tex]\large\qquad\qquad\qquad\tt MARCH/14/2025 [/tex]