Question

If a 10 kg mass is dropped from a height of 20 meters, what will its velocity be just before hitting the
ground? (Ignore air resistance and assume g=9.8 m/s2). use conservation of energy equation

Answer 1

To find the velocity of a 10 kg mass just before it hits the ground, we can use the principle of conservation of energy. According to this principle, the potential energy lost by the mass as it falls is converted into kinetic energy.

1. **Potential Energy (PE) at the height (h)**:
\[
PE = mgh
\]
where:
- \( m = 10 \, \text{kg} \) (mass),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 20 \, \text{m} \) (height).

Substituting the values:
\[
PE = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 20 \, \text{m} = 1960 \, \text{J}
\]

2. **Kinetic Energy (KE) just before hitting the ground**:
The kinetic energy just before impact is given by:
\[
KE = \frac{1}{2} mv^2
\]
At the point of impact, all the potential energy is converted to kinetic energy, so:
\[
PE = KE
\]
Therefore:
\[
1960 \, \text{J} = \frac{1}{2} \times 10 \, \text{kg} \times v^2
\]

3. **Solving for velocity (v)**:
\[
1960 = 5v^2
\]
\[
v^2 = \frac{1960}{5} = 392
\]
\[
v = \sqrt{392} \approx 19.8 \, \text{m/s}
\]

Thus, the velocity of the mass just before hitting the ground is approximately **19.8 m/s**.