To find the speed of a ball just before it hits the ground after being dropped from a height of 12 feet, we can use the formula derived from the equations of motion under constant acceleration:
\[ v^2 = u^2 + 2gh \]
Where:
- \( v \) is the final velocity (just before impact),
- \( u \) is the initial velocity (0 m/s since it is dropped),
- \( g \) is the acceleration due to gravity (9.8 m/s²),
- \( h \) is the height from which the ball is dropped.
First, we need to convert the height from feet to meters. Given that \( 1 \text{ ft} = 0.30 \text{ m} \):
\[ h = 12 \text{ ft} \times 0.30 \text{ m/ft} = 3.6 \text{ m} \]
Now we can plug the values into the equation:
\[ v^2 = 0 + 2 \cdot 9.8 \text{ m/s}² \cdot 3.6 \text{ m} \]
Calculating the right side:
\[ v^2 = 2 \cdot 9.8 \cdot 3.6 \] \[ v^2 = 70.56 \]
Now taking the square root to find \( v \):
\[ v = \sqrt{70.56} \approx 8.4 \text{ m/s} \]
Thus, the speed of the ball just before it hits the ground is approximately 8.4 m/s, which corresponds to the second option in your list.
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