1. **Problem Statement:** A projectile is thrown towards a basket located 75 m away horizontally and 7.4 m high vertically. We need to determine mathematically if the object will successfully land in the basket. 2. **Given Data:** - Horizontal distance, $x = 75$ m - Height of basket, $y = 7.4$ m - Initial velocity from graph approximately $v_0 = 15$ m/s at $t=0$ - Time at minimum velocity point $t = 2.65$ s 3. **Assumptions and Formulas:** - Assume projectile motion under gravity $g = 9.8$ m/s$^2$ - Horizontal velocity $v_x$ is constant (no air resistance) - Vertical velocity changes due to gravity - Horizontal distance formula: $$x = v_x t$$ - Vertical position formula: $$y = v_{y0} t - \frac{1}{2} g t^2$$ 4. **Find horizontal velocity $v_x$:** From the graph, initial velocity is about 15 m/s, and velocity decreases to 0 at $t=2.65$ s, indicating vertical velocity component. Assuming initial velocity is vertical component $v_{y0} = 15$ m/s. 5. **Calculate time to reach basket horizontally:** Assuming horizontal velocity $v_x$ is constant and unknown, time to reach basket horizontally is $$t = \frac{x}{v_x}$$ 6. **Calculate vertical position at time $t$:** Vertical position at time $t$ is $$y = v_{y0} t - \frac{1}{2} g t^2$$ 7. **Check if vertical position equals basket height:** We want to find $t$ such that horizontal distance is 75 m: $$t = \frac{75}{v_x}$$ Substitute into vertical position: $$y = 15 t - 4.9 t^2$$ Set $y = 7.4$ m: $$7.4 = 15 t - 4.9 t^2$$ Rearranged: $$4.9 t^2 - 15 t + 7.4 = 0$$ 8. **Solve quadratic equation:** Using quadratic formula: $$t = \frac{15 \pm \sqrt{15^2 - 4 \times 4.9 \times 7.4}}{2 \times 4.9}$$ Calculate discriminant: $$15^2 - 4 \times 4.9 \times 7.4 = 225 - 144.96 = 80.04$$ Calculate roots: $$t = \frac{15 \pm 8.95}{9.8}$$ Two solutions: - $$t_1 = \frac{15 - 8.95}{9.8} = 0.62 \text{ s}$$ - $$t_2 = \frac{15 + 8.95}{9.8} = 2.44 \text{ s}$$ 9. **Find horizontal velocity $v_x$ for each time:** $$v_x = \frac{75}{t}$$ - For $t_1 = 0.62$ s, $$v_x = \frac{75}{0.62} = 121.0 \text{ m/s}$$ (unrealistic) - For $t_2 = 2.44$ s, $$v_x = \frac{75}{2.44} = 30.7 \text{ m/s}$$ 10. **Compare with initial velocity:** Initial velocity magnitude from graph is about 15 m/s, but horizontal velocity needed is 30.7 m/s, which is higher. **Conclusion:** The object cannot reach the basket at 75 m distance and 7.4 m height with the given initial velocity of 15 m/s. Hence, the attempt to throw the object into the basket will not be successful mathematically.