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!

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.