1. **Problem statement:** A soccer player kicks a ball at an initial velocity at an angle of 41° above the horizontal. The ball passes over a front wall 30.7 m away horizontally at 1.8 s and lands on a flat roof 7.6 m high. We want to analyze the projectile motion to find relevant parameters. 2. **Relevant formulas:** For projectile motion, horizontal and vertical motions are independent. - Horizontal position: $$x = v_0 \cos(\theta) t$$ - Vertical position: $$y = v_0 \sin(\theta) t - \frac{1}{2} g t^2$$ where $v_0$ is initial velocity, $\theta = 41^\circ$, $g = 9.8$ m/s², $t$ is time. 3. **Given data:** - $\theta = 41^\circ$ - Horizontal distance to wall: $x = 30.7$ m - Height of roof: $y = 7.6$ m - Time to pass wall: $t = 1.8$ s 4. **Find initial velocity $v_0$ using horizontal motion at $t=1.8$ s:** $$x = v_0 \cos(41^\circ) \times 1.8$$ $$v_0 = \frac{30.7}{1.8 \cos(41^\circ)}$$ Calculate $\cos(41^\circ) \approx 0.7547$: $$v_0 = \frac{30.7}{1.8 \times 0.7547} = \frac{30.7}{1.3585} \approx 22.6$$ 5. **Check vertical position at $t=1.8$ s:** $$y = v_0 \sin(41^\circ) \times 1.8 - \frac{1}{2} \times 9.8 \times (1.8)^2$$ Calculate $\sin(41^\circ) \approx 0.6561$: $$y = 22.6 \times 0.6561 \times 1.8 - 4.9 \times 3.24$$ $$y = 26.7 - 15.9 = 10.8$$ This is the height of the ball when it passes the wall, which is above the roof height 7.6 m, so the ball passes over the wall. 6. **Find time $t_r$ when ball lands on roof at height 7.6 m:** Set vertical position equal to roof height: $$7.6 = 22.6 \times 0.6561 \times t_r - 4.9 t_r^2$$ Simplify: $$7.6 = 14.83 t_r - 4.9 t_r^2$$ Rearranged: $$4.9 t_r^2 - 14.83 t_r + 7.6 = 0$$ 7. **Solve quadratic equation for $t_r$:** $$t_r = \frac{14.83 \pm \sqrt{14.83^2 - 4 \times 4.9 \times 7.6}}{2 \times 4.9}$$ Calculate discriminant: $$14.83^2 = 219.9, \quad 4 \times 4.9 \times 7.6 = 149.0$$ $$\sqrt{219.9 - 149.0} = \sqrt{70.9} = 8.42$$ Calculate roots: $$t_r = \frac{14.83 \pm 8.42}{9.8}$$ Two solutions: - $$t_r = \frac{14.83 + 8.42}{9.8} = \frac{23.25}{9.8} = 2.37$$ - $$t_r = \frac{14.83 - 8.42}{9.8} = \frac{6.41}{9.8} = 0.65$$ 8. **Interpretation:** The ball reaches height 7.6 m twice: ascending at 0.65 s and descending at 2.37 s. Since it passes the wall at 1.8 s, the landing on the roof is at $t_r = 2.37$ s. 9. **Find horizontal distance at landing:** $$x_r = 22.6 \times \cos(41^\circ) \times 2.37 = 22.6 \times 0.7547 \times 2.37 = 40.4$$ **Final answers:** - Initial velocity $v_0 \approx 22.6$ m/s - Time to land on roof $t_r \approx 2.37$ s - Horizontal distance to roof landing $x_r \approx 40.4$ m These results match the problem data and confirm the ball passes over the wall and lands on the roof.