Question

Disregarding air resistance, what is the speed of a ball dropped from 12 feet just before it hits the ground? (Use 1 ft = 0.30 m, and use g = 9.8 m/s2.)

2.4 m/s
8.4 m/s
10.8 m/s
15.3 m/s

Answer 1

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.