A basketball player shoots toward a basket 7 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor at an angle of 60 degrees
 above the horizontal, what must the initial speed be if it were to go through the basket?

We can solve this problem using kinematic equations for projectile motion. Here's the step-by-step process:

### Step 1: Define the known variables.
- Horizontal distance to the basket, \( x = 7 \, \text{m} \)
- Height of the basket above the floor, \( y_b = 3.0 \, \text{m} \)
- Height of the ball when released, \( y_0 = 1.8 \, \text{m} \)
- Angle of projection, \( \theta = 60^\circ \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

We need to find the initial speed \( v_0 \) such that the ball passes through the basket.

### Step 2: Set up the kinematic equations.

The projectile motion equations for horizontal and vertical components are:

#### Horizontal motion:
\[
x = v_0 \cos(\theta) \cdot t
\]
Where \( t \) is the time of flight.

#### Vertical motion:
\[
y_b = y_0 + v_0 \sin(\theta) \cdot t - \frac{1}{2} g t^2
\]
Where \( y_b \) is the height of the basket.

### Step 3: Solve for time \( t \) from the horizontal equation.
From the horizontal motion equation, we solve for \( t \):

\[
t = \frac{x}{v_0 \cos(\theta)}
\]

Substitute this into the vertical motion equation.

### Step 4: Substitute \( t \) into the vertical equation.

Substitute \( t = \frac{7}{v_0 \cos(60^\circ)} \) into the vertical equation:

\[
3.0 = 1.8 + v_0 \sin(60^\circ) \cdot \frac{7}{v_0 \cos(60^\circ)} - \frac{1}{2} \cdot 9.8 \cdot \left( \frac{7}{v_0 \cos(60^\circ)} \right)^2
\]

### Step 5: Simplify and solve for \( v_0 \).

First, calculate the trigonometric values:
- \( \cos(60^\circ) = 0.5 \)
- \( \sin(60^\circ) \approx 0.866 \)

Substituting these into the equation:

\[
3.0 = 1.8 + 0.866 \cdot \frac{7}{v_0 \cdot 0.5} - \frac{1}{2} \cdot 9.8 \cdot \left( \frac{7}{v_0 \cdot 0.5} \right)^2
\]

Simplifying this leads to a quadratic equation in terms of \( v_0 \). Let's now solve it.The initial speed \( v_0 \) required for the ball to go through the basket is approximately \( 9.38 \, \text{m/s} \). Since speed cannot be negative, the valid solution is:

\[
v_0 = 9.38 \, \text{m/s}
\]

Would you like more details on any part of this solution, or do you have further questions?

Here are 5 related questions to think about:
1. How would the initial speed change if the distance to the basket were increased?
2. What if the angle of projection were 45 degrees instead of 60 degrees?
3. How does air resistance affect the actual motion of the basketball?
4. What would be the time of flight for this basketball shot?
5. How would the trajectory differ if the player were to shoot the ball from a higher initial position?

**Tip:** In projectile motion, the horizontal and vertical motions are independent of each other, which simplifies the calculation process by treating them separately.

A basketball player shoots toward a basket 7 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor at an angle of 60 degrees above the horizontal, what must the initial speed be if it were to go through the basket?

Find the initial speed needed for a basketball shot from 7 meters away at a 60° angle to go through a basket 3.0 m above the floor. This detailed solution covers projectile motion, kinematic equations, and trigonometry. Ideal for high school students in Grades 10-12.

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