91Ó°ÊÓ

In the context of projectile motion, initial velocity is the speed at which an object is launched. In our basketball exercise, it's the speed imparted to the ball by the player. The initial velocity determines how high and how far the ball will travel.
The ball is thrown up, suggesting that its initial velocity is directed vertically. Understanding this is crucial because it informs us about how long the ball stays airborne and how high it will go.
For our problem, we recognize that initial velocity (\(v_0\)) must be calculated such that it allows the ball to stay in motion for at least 4.8 seconds without touching the roof. This can be derived using known physics formulas involving time and gravitational pull.

Acceleration due to gravity is a constant force experienced by all objects when in free fall near the surface of the Earth. It is symbolized by \(g\) and has a standard value of \(9.81 \, \text{m/s}^2\).
In this context, gravity works against the motion of the ball, slowing it down as it ascends and accelerating it on its descent.
A key part of solving our basketball exercise involves accounting for gravity's impact on the ball's motion. It helps us calculate how long it takes for the ball to reach its peak height (where its upward speed is zero) and then return to the player's hand.

The time of flight is the total time that the projectile, in this case, the basketball, stays in the air.
To determine if the basketball stays in the air for at least 4.8 seconds, we use the principle that the time to ascend to the maximum height is equal to the time to descend back to the release point.
To calculate this, we use the formula: \[T = 2 \times \frac{v_0}{g}\]This equation calculates the total time of flight based on the initial velocity \(v_0\) and the constant acceleration due to gravity \(g\).
Thus, solving for \(T\) determines whether the player successfully runs out the clock.

In projectile motion, the maximum height is the highest point that the projectile reaches. In our scenario, it's crucial to ensure the basketball doesn't exceed 17.2 meters, avoiding a foul by touching the roof.
The maximum height \(h\) can be calculated using: \[h = h_0 + \frac{v_0^2}{2g}\]where \(h_0\) is the initial height of the ball and \(g\) is the gravitational acceleration.
By rearranging this formula to solve for \(v_0\), we can ensure that the maximum height remains below the predefined limit, while also checking the criteria of the flight time.This ensures both safety and strategic play, as the ball needs to remain in air without breaching structural or gameplay rules.