A student drops a basketball into a pool from a height
When a student drops a basketball into a pool from a height, several factors come into play:
1. Gravitational force: As soon as the basketball is released from the student's hand, it is subject to the force of gravity pulling it towards the ground.
2. Air resistance: As the basketball falls through the air, it experiences air resistance. This force opposes the motion of the basketball and gradually reduces its speed.
3. Buoyancy: Once the basketball enters the water, it displaces an equivalent amount of water. According to Archimedes' principle, the basketball experiences an upward buoyant force equal to the weight of the water it displaces. This buoyant force acts in the opposite direction to gravity and reduces the net downward force on the basketball.
4. Drag in water: Unlike air, water is much denser and presents a higher resistance to motion. As the basketball descends through the water, it encounters drag, which further reduces its speed.
Overall, as the basketball drops into the pool from a height, its speed gradually decreases due to the opposing forces of gravity, air resistance, buoyancy, and water drag. The exact behavior of the basketball's descent depends on the specific conditions, such as the height from which it is dropped and the characteristics of the pool and the ball itself.
To provide a step-by-step answer, I need to know the specific height from which the student dropped the basketball.
To calculate the time it takes for a basketball to hit the water when dropped from a certain height, you can use the equation of motion. The equation for calculating the time it takes for an object to fall freely can be represented as:
t = √(2h/g)
Where:
t = time taken (in seconds)
h = height (in meters)
g = acceleration due to gravity (approximately 9.8 m/s² on Earth)
To find the time it takes for the basketball to hit the water, you'll need to know the height from which it was dropped. Let's assume the height is 10 meters.
Plugging this value into the equation, we get:
t = √(2*10/9.8)
Simplifying further:
t = √(20/9.8)
t = √2.04
t ≈ 1.43 seconds
Therefore, it would take approximately 1.43 seconds for the basketball to hit the water when dropped from a height of 10 meters.