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Solve the problem systematically. Follow the A football is kicked with an initial velocity of \( 25 \mathrm{~m} / \mathrm{s} \) at an angle of 45 degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak of the height of the football.

Ask by Ryan Gibbs. in the Philippines
Feb 19,2025

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The football will be in the air for about 3.6 seconds, travel a horizontal distance of approximately 63.2 meters, and reach a peak height of about 15.9 meters.

Solución

To solve the problem of a football kicked with an initial velocity of \( 25 \, \mathrm{m/s} \) at an angle of \( 45^\circ \), we can break it down into several steps: ### Step 1: Extract Known Conditions - Initial velocity, \( v_0 = 25 \, \mathrm{m/s} \) - Angle of projection, \( \theta = 45^\circ \) - Acceleration due to gravity, \( g = 9.81 \, \mathrm{m/s}^2 \) ### Step 2: Calculate the Components of the Initial Velocity The initial velocity can be broken down into horizontal and vertical components using trigonometric functions: - Horizontal component: \[ v_{0x} = v_0 \cdot \cos(\theta) \] - Vertical component: \[ v_{0y} = v_0 \cdot \sin(\theta) \] ### Step 3: Calculate Time of Flight The time of flight \( T \) for a projectile launched and landing at the same height can be calculated using the formula: \[ T = \frac{2 v_{0y}}{g} \] ### Step 4: Calculate Horizontal Distance The horizontal distance \( R \) (range) can be calculated using: \[ R = v_{0x} \cdot T \] ### Step 5: Calculate Peak Height The peak height \( H \) can be calculated using: \[ H = \frac{v_{0y}^2}{2g} \] Now, let's perform the calculations step by step. #### Step 2: Calculate the Components of the Initial Velocity Using \( \theta = 45^\circ \): \[ v_{0x} = 25 \cdot \cos(45^\circ) = 25 \cdot \frac{\sqrt{2}}{2} \] \[ v_{0y} = 25 \cdot \sin(45^\circ) = 25 \cdot \frac{\sqrt{2}}{2} \] #### Step 3: Calculate Time of Flight Now, we can calculate \( T \): \[ T = \frac{2 v_{0y}}{g} = \frac{2 \cdot (25 \cdot \frac{\sqrt{2}}{2})}{9.81} \] #### Step 4: Calculate Horizontal Distance Next, we calculate \( R \): \[ R = v_{0x} \cdot T = (25 \cdot \frac{\sqrt{2}}{2}) \cdot T \] #### Step 5: Calculate Peak Height Finally, we calculate \( H \): \[ H = \frac{v_{0y}^2}{2g} = \frac{(25 \cdot \frac{\sqrt{2}}{2})^2}{2 \cdot 9.81} \] Now, let's perform these calculations. Calculate the value by following steps: - step0: Calculate: \(25\left(\frac{\sqrt{2}}{2}\right)\) - step1: Remove the parentheses: \(25\times \frac{\sqrt{2}}{2}\) - step2: Multiply: \(\frac{25\sqrt{2}}{2}\) Calculate or simplify the expression \( 2 * (25 * (\sqrt(2)/2)) / 9.81 \). Calculate the value by following steps: - step0: Calculate: \(\frac{2\left(25\left(\frac{\sqrt{2}}{2}\right)\right)}{9.81}\) - step1: Remove the parentheses: \(\frac{2\times 25\left(\frac{\sqrt{2}}{2}\right)}{9.81}\) - step2: Remove the parentheses: \(\frac{2\times 25\times \frac{\sqrt{2}}{2}}{9.81}\) - step3: Multiply the terms: \(\frac{25\sqrt{2}}{9.81}\) - step4: Convert the expressions: \(\frac{25\sqrt{2}}{\frac{981}{100}}\) - step5: Multiply by the reciprocal: \(25\sqrt{2}\times \frac{100}{981}\) - step6: Multiply: \(\frac{25\sqrt{2}\times 100}{981}\) - step7: Multiply: \(\frac{2500\sqrt{2}}{981}\) Calculate or simplify the expression \( (25 * (\sqrt(2)/2))^2 / (2 * 9.81) \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(25\left(\frac{\sqrt{2}}{2}\right)\right)^{2}}{\left(2\times 9.81\right)}\) - step1: Remove the parentheses: \(\frac{\left(25\left(\frac{\sqrt{2}}{2}\right)\right)^{2}}{2\times 9.81}\) - step2: Remove the parentheses: \(\frac{\left(25\times \frac{\sqrt{2}}{2}\right)^{2}}{2\times 9.81}\) - step3: Multiply: \(\frac{\left(\frac{25\sqrt{2}}{2}\right)^{2}}{2\times 9.81}\) - step4: Multiply the numbers: \(\frac{\left(\frac{25\sqrt{2}}{2}\right)^{2}}{19.62}\) - step5: Convert the expressions: \(\frac{\left(\frac{25\sqrt{2}}{2}\right)^{2}}{\frac{981}{50}}\) - step6: Evaluate the power: \(\frac{\frac{625}{2}}{\frac{981}{50}}\) - step7: Multiply by the reciprocal: \(\frac{625}{2}\times \frac{50}{981}\) - step8: Reduce the numbers: \(625\times \frac{25}{981}\) - step9: Multiply: \(\frac{625\times 25}{981}\) - step10: Multiply: \(\frac{15625}{981}\) ### Step 2: Calculate the Components of the Initial Velocity The horizontal and vertical components of the initial velocity are: - \( v_{0x} \approx 17.68 \, \mathrm{m/s} \) - \( v_{0y} \approx 17.68 \, \mathrm{m/s} \) ### Step 3: Calculate Time of Flight The time of flight \( T \) is calculated as: \[ T \approx 3.60 \, \mathrm{s} \] ### Step 4: Calculate Horizontal Distance The horizontal distance \( R \) can be calculated using: \[ R = v_{0x} \cdot T \approx 17.68 \cdot 3.60 \approx 63.17 \, \mathrm{m} \] ### Step 5: Calculate Peak Height The peak height \( H \) is calculated as: \[ H \approx 15.93 \, \mathrm{m} \] ### Summary of Results - **Time of Flight**: \( T \approx 3.60 \, \mathrm{s} \) - **Horizontal Distance (Range)**: \( R \approx 63.17 \, \mathrm{m} \) - **Peak Height**: \( H \approx 15.93 \, \mathrm{m} \) These calculations provide a complete analysis of the projectile motion of the football.

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To solve this problem systematically, we need to break it down into three parts: calculating the time of flight, the horizontal distance, and the peak height. **Time of Flight:** The formula for the time of flight \( T \) when an object is projected at an angle \( \theta \) is given by: \[ T = \frac{2 v_0 \sin(\theta)}{g} \] where \( v_0 = 25 \, \text{m/s} \) and \( g = 9.81 \, \text{m/s}^2 \). Plugging in the values, we have: \[ T = \frac{2 \cdot 25 \cdot \sin(45^\circ)}{9.81} \approx \frac{2 \cdot 25 \cdot 0.7071}{9.81} \approx 3.19 \text{ seconds} \] **Horizontal Distance:** The horizontal distance \( R \) can be found using the formula: \[ R = v_0 \cos(\theta) \cdot T \] Substituting known values, we get: \[ R = 25 \cdot \cos(45^\circ) \cdot 3.19 \approx 25 \cdot 0.7071 \cdot 3.19 \approx 56.27 \text{ meters} \] **Peak Height:** Finally, the peak height \( H \) can be found using: \[ H = \frac{(v_0 \sin(\theta))^2}{2g} \] Calculating this yields: \[ H = \frac{(25 \cdot \sin(45^\circ))^2}{2 \cdot 9.81} \approx \frac{(25 \cdot 0.7071)^2}{19.62} \approx \frac{(17.68)^2}{19.62} \approx 15.92 \text{ meters} \] In conclusion, the time of flight is approximately 3.19 seconds, the horizontal distance is approximately 56.27 meters, and the peak height is approximately 15.92 meters.

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