Answer:
To determine whether the ball will reach the goal post, we need to analyze the projectile motion of the football. We'll break the motion into horizontal and vertical components and calculate the trajectory of the ball to see if it reaches the height of the goal post at the distance of 40 yards.
### Given Data:
- Distance to goalpost \(d = 40 \text{ yards} = 120 \text{ feet}\)
- Height of goalpost \(h = 10 \text{ feet}\)
- Initial speed \(v_0 = 64 \text{ ft/s}\)
- Launch angle \(\theta = 45^\circ\)
- Acceleration due to gravity \(g = 32.2 \text{ ft/s}^2\)
### Horizontal Motion:
The horizontal component of the initial velocity \(v_{0x}\) is given by:
\[ v_{0x} = v_0 \cos(\theta) = 64 \cos(45^\circ) = 64 \cdot \frac{\sqrt{2}}{2} = 32\sqrt{2} \text{ ft/s} \]
The horizontal distance traveled \(d_x\) after time \(t\) is:
\[ d_x = v_{0x} t \]
To reach the goalpost 120 feet away:
\[ 120 = 32\sqrt{2} \cdot t \]
\[ t = \frac{120}{32\sqrt{2}} \]
\[ t = \frac{120}{32 \cdot 1.414} \]
\[ t \approx 2.65 \text{ seconds} \]
### Vertical Motion:
The vertical component of the initial velocity \(v_{0y}\) is given by:
\[ v_{0y} = v_0 \sin(\theta) = 64 \sin(45^\circ) = 64 \cdot \frac{\sqrt{2}}{2} = 32\sqrt{2} \text{ ft/s} \]
The vertical position \(y\) after time \(t\) is given by:
\[ y = v_{0y} t - \frac{1}{2} g t^2 \]
\[ y = 32\sqrt{2} \cdot t - \frac{1}{2} \cdot 32.2 \cdot t^2 \]
Substituting \(t \approx 2.65 \text{ seconds}\):
\[ y = 32\sqrt{2} \cdot 2.65 - \frac{1}{2} \cdot 32.2 \cdot (2.65)^2 \]
\[ y = 32 \cdot 1.414 \cdot 2.65 - 16.1 \cdot 7.0225 \]
\[ y \approx 120 \text{ ft} - 113.6645 \text{ ft} \]
\[ y \approx 6.3355 \text{ ft} \]
### Conclusion:
The height of the ball when it reaches the goalpost 120 feet away is approximately 6.34 feet, which is less than the height of the goalpost (10 feet). Therefore, the ball will not reach the goal post.
