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To solve this problem, we need to find the initial speed \(v_0\) at which the basketball player must throw the ball at an angle of \(\theta_0 = 45^\circ\) to successfully make the shot.

Given dataTo solve this problem, we need to find the initial speed \(v_0\) at which the basketball player must throw the ball at an angle of \(\theta_0 = 45^\circ\) to successfully make the shot.

Given 
- Horizontal distances: \(d_1 = 1.0 \, \text{ft}\) (initial horizontal distance), \(d_2 = 14 \, \text{ft}\) (total horizontal distance).
- Heights: \(h_1 = 7.8 \, \text{ft}\) (initial height), \(h_2 = 10 \, \text{ft}\) (final height of the hoop).
- Angle of projection: \(\theta_0 = 45^\circ\).

### Step-by-step solution:

1. **Determine the total horizontal distance**:  
   Total horizontal distance is \(D = d_1 + d_2 = 1.0 \, \text{ft} + 14 \, \text{ft} = 15.0 \, \text{ft}\).

2. **Equations of motion**:
   We can use the equations for projectile motion. The horizontal and vertical displacements are described by:

   - Horizontal motion:
     \[
     x(t) = v_0 \cos(\theta_0) t
     \]
     where \(x(t) = D\) is the total horizontal distance, and \(t\) is the time of flight.
   
   - Vertical motion:
     \[
     y(t) = h_1 + v_0 \sin(\theta_0) t - \frac{1}{2} g t^2
     \]
     where \(y(t) = h_2\) is the final height of the ball, and \(g = 32.2 \, \text{ft/s}^2\) is the acceleration due to gravity.

3. **Find the time of flight \(t\)**:
   From the horizontal motion equation, solve for \(t\):
   \[
   t = \frac{D}{v_0 \cos(\theta_0)}
   \]
   Since \(\theta_0 = 45^\circ\), \(\cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2}\), so this simplifies to:
   \[
   t = \frac{D}{v_0 \cdot \frac{\sqrt{2}}{2}} = \frac{2D}{v_0 \sqrt{2}}
   \]

4. **Substitute into the vertical motion equation**:
   Now substitute \(t\) into the vertical motion equation:
   \[
   h_2 = h_1 + v_0 \sin(45^\circ) t - \frac{1}{2} g t^2
   \]
   Again, \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\), so:
   \[
   h_2 = h_1 + v_0 \cdot \frac{\sqrt{2}}{2} \cdot \frac{2D}{v_0 \sqrt{2}} - \frac{1}{2} g \left(\frac{2D}{v_0 \sqrt{2}}\right)^2
   \]
   Simplifying:
   \[
   h_2 = h_1 + D - \frac{g D^2}{2 v_0^2}
   \]

5. **Solve for \(v_0\)**:
   Rearranging the equation to solve for \(v_0\):
   \[
   v_0^2 = \frac{g D^2}{2 (h_1 + D - h_2)}
   \]
   Substituting the known values:
   \[
   v_0^2 = \frac{32.2 \times (15)^2}{2 (7.8 + 15 - 10)}
   \]
   Simplifying:
   \[
   v_0^2 = \frac{32.2 \times 225}{2 (12.8)}
   \]
   \[
   v_0^2 = \frac{32.2 \times 225}{25.6}
   \]
   \[
   v_0^2 \approx 282.42
   \]
   Taking the square root:
   \[
   v_0 \approx 16.81 \, \text{ft/s}
   \]

Thus, the initial speed \(v_0 \approx 16.8 \, \text{ft/s}\).

Let me know if you would like further clarification, details, or if you have any questions!

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**Related questions:**
1. How does air resistance affect the trajectory of a projectile?
2. Can you derive the time of flight for a projectile in general terms?
3. How would the required initial speed change if the angle was different from \(45^\circ\)?
4. What is the optimal angle to achieve the maximum range in projectile motion?
5. How do we account for the Earth's curvature in long-distance projectile motion?

**Tip:** Always check units when solving physics problems to ensure consistency across calculations.

At what initial speed must the basketball player in the figure throw the ball, at angle θ₀ = 45° above the horizontal, to make the foul shot? The horizontal distances are d₁ = 1.0 ft and d₂ = 14 ft, and the heights are h₁ = 7.8 ft and h₂ = 10 ft.

Learn how to calculate the initial speed for a basketball shot at a 45-degree angle. This problem involves projectile motion, horizontal and vertical displacements, and kinematic equations. Given the distances and heights, the solution determines the initial speed required for the shot. Suitable for high school students.

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