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).
ground? (Ignore air resistance and assume g=9.8 m/s2).
Answers
GPT-4o mini
To calculate the velocity of the mass just before hitting the ground, we can use the principle of conservation of energy or the kinematic equation. Here, I'll use the kinematic equation for an object in free fall:
\[
v^2 = u^2 + 2gh
\]
Where:
- \( v \) is the final velocity (just before hitting the ground),
- \( u \) is the initial velocity (which is 0 m/s since the mass is dropped),
- \( g \) is the acceleration due to gravity (9.8 m/s²),
- \( h \) is the height from which it is dropped (20 m).
Substituting the values into the equation:
\[
v^2 = 0 + 2 \times 9.8 \, \text{m/s}^2 \times 20 \, \text{m}
\]
\[
v^2 = 2 \times 9.8 \times 20
\]
\[
v^2 = 392
\]
Now, take the square root of both sides to find \( v \):
\[
v = \sqrt{392} \approx 19.8 \, \text{m/s}
\]
Thus, the velocity of the mass just before hitting the ground will be approximately **19.8 m/s**.
\[
v^2 = u^2 + 2gh
\]
Where:
- \( v \) is the final velocity (just before hitting the ground),
- \( u \) is the initial velocity (which is 0 m/s since the mass is dropped),
- \( g \) is the acceleration due to gravity (9.8 m/s²),
- \( h \) is the height from which it is dropped (20 m).
Substituting the values into the equation:
\[
v^2 = 0 + 2 \times 9.8 \, \text{m/s}^2 \times 20 \, \text{m}
\]
\[
v^2 = 2 \times 9.8 \times 20
\]
\[
v^2 = 392
\]
Now, take the square root of both sides to find \( v \):
\[
v = \sqrt{392} \approx 19.8 \, \text{m/s}
\]
Thus, the velocity of the mass just before hitting the ground will be approximately **19.8 m/s**.