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

Answer: 8.4 m/s

Explanation:

This described situation is related to vertical motion (free fall), with constant acceleration (acceleration due gravity), and in this case, the following formula will be useful:

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (1)

Where:

$V_{f}$ Is the final velocity of the ball

$V_{o}=0$ Is the initial velocity of the ball (it was dropped)

$a=9.8 m/s^{2}$ is the acceleration due gravity

$d=12 ft=3.6 m$ is the distance at which the ball hits the ground (taking into account $1 ft=0.30 m$, $12 ft \frac{0.30 m}{1 ft}=3.6 m$)

${V_{f}}^{2}=0+2(9.8 m/s^{2})(3.6 m)$ (2)

$V_{f}=\sqrt{2(9.8 m/s^{2})(3.6 m)}$ (3)

Finally:

$V_{f}=8.4 m/s$

Answer 2

Answer:

B - 8.4 m/ is the correct answer.