Soccer Ball Speed 1C42Ab

1. **State the problem:** A soccer ball is kicked at an angle of $45^\circ$ from the ground and hits a crossbar 10 m away horizontally and 2.5 m above the ground. We need to find the initial speed $v_0$ of the ball. 2. **Known values:** - Horizontal distance: $\Delta x = 10$ m - Vertical height at crossbar: $y = 2.5$ m - Launch angle: $\theta = 45^\circ$ - Acceleration due to gravity: $g = 10$ m/s$^2$ 3. **Formulas used:** - Horizontal motion: $\Delta x = v_0 \cos\theta \cdot t$ - Vertical motion: $y = v_0 \sin\theta \cdot t - \frac{1}{2} g t^2$ 4. **Step 1: Express time $t$ from horizontal motion:** $$ t = \frac{\Delta x}{v_0 \cos\theta} = \frac{10}{v_0 \cos 45^\circ} $$ Since $\cos 45^\circ = \frac{\sqrt{2}}{2}$, this becomes: $$ t = \frac{10}{v_0 \cdot \frac{\sqrt{2}}{2}} = \frac{10 \times 2}{v_0 \sqrt{2}} = \frac{20}{v_0 \sqrt{2}} $$ 5. **Step 2: Substitute $t$ into vertical motion equation:** $$ 2.5 = v_0 \sin 45^\circ \cdot t - \frac{1}{2} \times 10 \times t^2 $$ Since $\sin 45^\circ = \frac{\sqrt{2}}{2}$, substitute $t$: $$ 2.5 = v_0 \cdot \frac{\sqrt{2}}{2} \cdot \frac{20}{v_0 \sqrt{2}} - 5 \left( \frac{20}{v_0 \sqrt{2}} \right)^2 $$ 6. **Step 3: Simplify the first term:** $$ v_0 \cdot \frac{\sqrt{2}}{2} \cdot \frac{20}{v_0 \sqrt{2}} = \frac{\sqrt{2}}{2} \cdot \frac{20}{\sqrt{2}} = \frac{20}{2} = 10 $$ 7. **Step 4: Simplify the second term:** $$ 5 \left( \frac{20}{v_0 \sqrt{2}} \right)^2 = 5 \cdot \frac{400}{v_0^2 \times 2} = 5 \cdot \frac{400}{2 v_0^2} = 5 \cdot \frac{200}{v_0^2} = \frac{1000}{v_0^2} $$ 8. **Step 5: Write the equation:** $$ 2.5 = 10 - \frac{1000}{v_0^2} $$ 9. **Step 6: Solve for $v_0^2$:** $$ 2.5 - 10 = - \frac{1000}{v_0^2} \Rightarrow -7.5 = - \frac{1000}{v_0^2} \Rightarrow 7.5 = \frac{1000}{v_0^2} $$ $$ v_0^2 = \frac{1000}{7.5} = 133.33 $$ 10. **Step 7: Calculate $v_0$:** $$ v_0 = \sqrt{133.33} \approx 11.55 \text{ m/s} $$ 11. **Final answer:** The initial speed is approximately $11.5$ m/s, which corresponds to option (C). **Answer: (C) 11.5 m/s**