Cannonball Motion 3Ae16A

1. **State the problem:** We need to find the position and velocity of a cannon ball shot with an initial velocity of $141.4$ m/s at an angle of $45^\circ$ after $5$ seconds. 2. **Formulas and important rules:** - Initial velocity components: $$v_{0x} = v_0 \cos \theta$$ $$v_{0y} = v_0 \sin \theta$$ - Position at time $t$: $$x = v_{0x} t$$ $$y = v_{0y} t - \frac{1}{2} g t^2$$ - Velocity components at time $t$: $$v_x = v_{0x}$$ $$v_y = v_{0y} - g t$$ where $g = 9.8$ m/s$^2$ is acceleration due to gravity. 3. **Calculate initial velocity components:** $$v_{0x} = 141.4 \times \cos 45^\circ = 141.4 \times \frac{\sqrt{2}}{2} = 141.4 \times 0.7071 = 100$$ $$v_{0y} = 141.4 \times \sin 45^\circ = 141.4 \times 0.7071 = 100$$ 4. **Calculate position at $t=5$ s:** $$x = 100 \times 5 = 500$$ $$y = 100 \times 5 - \frac{1}{2} \times 9.8 \times 5^2 = 500 - \frac{1}{2} \times 9.8 \times 25 = 500 - 122.5 = 377.5$$ 5. **Calculate velocity components at $t=5$ s:** $$v_x = 100$$ $$v_y = 100 - 9.8 \times 5 = 100 - 49 = 51$$ 6. **Final velocity magnitude and direction:** $$v = \sqrt{v_x^2 + v_y^2} = \sqrt{100^2 + 51^2} = \sqrt{10000 + 2601} = \sqrt{12601} \approx 112.25$$ $$\theta = \tan^{-1} \left( \frac{v_y}{v_x} \right) = \tan^{-1} \left( \frac{51}{100} \right) \approx 27^\circ$$ **Answer:** - Position at $5$ s: $(x, y) = (500, 377.5)$ m - Velocity at $5$ s: magnitude $\approx 112.25$ m/s, angle $\approx 27^\circ$ above horizontal