To find the range of the basketball, we need to determine the horizontal distance it traveled before hitting the ground.

We are given:

• Horizontal velocity of the ball, $v_x = 8 \, \text{m/s}$
• Time the ball was in the air, $t = 0.64 \, \text{seconds}$

The horizontal range of the ball is found using the formula:

$\text{Range} = v_x \times t$

Substituting the values:

$\text{Range} = 8 \, \text{m/s} \times 0.64 \, \text{s} = 5.12 \, \text{m}$

Rounding to the nearest whole number:

$\text{Range} \approx 5 \, \text{meters}$

Thus, the range of the basketball is approximately 5 meters.

Would you like further details or have any questions about the solution?

Related Questions:

1. How can we calculate the initial height of the ball if gravity is the only vertical force?
2. How does the horizontal velocity change if the ball were thrown at a different angle?
3. What would happen to the range if the time of flight doubled?
4. How can air resistance affect the horizontal range in real scenarios?
5. How can the vertical velocity be calculated if the initial height is known?

Tip:

In projectile motion, horizontal and vertical motions are independent. Horizontal velocity remains constant (neglecting air resistance), while vertical motion is affected by gravity.