A basketball referee tosses the ball straight up for the starting tipoff. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

To find the velocity with which a basketball player must leave the ground to rise 1.25 meters, we can use the following kinematic equation:

\[
v^2 = u^2 + 2as
\]

Where:
- \( v \) is the final velocity at the peak of the jump, which is \( 0 \) m/s (since the player stops rising at the peak),
- \( u \) is the initial velocity, which we need to find,
- \( a \) is the acceleration due to gravity, which is \( -9.8 \, \text{m/s}^2 \) (negative because it acts downward),
- \( s \) is the displacement, which is \( 1.25 \, \text{m} \).

Substituting the values into the equation:

\[
0 = u^2 + 2(-9.8)(1.25)
\]

Simplifying this:

\[
0 = u^2 - 24.5
\]

\[
u^2 = 24.5
\]

\[
u = \sqrt{24.5} \approx 4.95 \, \text{m/s}
\]

So, the basketball player must leave the ground with an initial velocity of approximately **4.95 m/s**.

Would you like any further details or have any questions?

Here are five related questions:
1. How long will it take for the player to reach the peak height?
2. What is the total time the player will be in the air during this jump?
3. How high would the player rise if they left the ground with a velocity of 6 m/s?
4. What would be the impact of a different gravitational acceleration on the required initial velocity?
5. How does the player's mass affect the initial velocity needed for the jump?

**Tip:** Always pay attention to the signs of acceleration and velocity when solving kinematic equations; they indicate the direction of motion.

Learn how to calculate the initial velocity required for a basketball player to reach a height of 1.25 meters. This problem involves kinematic equations and projectile motion concepts. Suitable for advanced high school students.

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