Apologies for the incorrect previous response. To calculate the height from which the basketball falls, we need to consider the vector components of the velocity.
Let's assume the basketball is dropped from rest, which means its initial velocity is zero. When it hits the ground, its final velocity is given as 7 m/s. Assuming negligible air resistance, we can use the equations of motion to find the height.
The equation we can use is:
v^2 = u^2 + 2as
Where:
- v is the final velocity (7 m/s)
- u is the initial velocity (0 m/s)
- a is the acceleration due to gravity (-9.8 m/s^2, taking negative sign as it acts opposite to the motion)
- s is the distance traveled (height)
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Substituting the values, we get:
s = (7^2 - 0^2) / (2 * -9.8)
s = 49 / -19.6
s ≈ -2.5 meters
The negative sign indicates that the height is below the starting point. Therefore, the correct answer is that the basketball was dropped from a height of approximately 2.5 meters below the starting point.