Question
What is the speed of an object dropped from a 56-meter height after it has traveled 2.5 meters downward?
Answers
GPT-4o mini
To calculate the speed of an object dropped from a height after it has traveled a certain distance downward, we can use the following physics formula derived from the equations of motion:
\[
v^2 = u^2 + 2a s
\]
Where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( s \) is the distance traveled.
In this case:
- The object is dropped, so the initial velocity \( u = 0 \, \text{m/s} \).
- The acceleration \( a = 9.81 \, \text{m/s}^2 \).
- The distance \( s = 2.5 \, \text{m} \).
Plugging in the values:
\[
v^2 = 0^2 + 2 \cdot 9.81 \cdot 2.5
\]
Calculating that:
\[
v^2 = 0 + 2 \cdot 9.81 \cdot 2.5 = 49.05
\]
Now, take the square root to find \( v \):
\[
v = \sqrt{49.05} \approx 7.0 \, \text{m/s}
\]
Therefore, the speed of the object after it has traveled 2.5 meters downward is approximately **7.0 m/s**.
\[
v^2 = u^2 + 2a s
\]
Where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( s \) is the distance traveled.
In this case:
- The object is dropped, so the initial velocity \( u = 0 \, \text{m/s} \).
- The acceleration \( a = 9.81 \, \text{m/s}^2 \).
- The distance \( s = 2.5 \, \text{m} \).
Plugging in the values:
\[
v^2 = 0^2 + 2 \cdot 9.81 \cdot 2.5
\]
Calculating that:
\[
v^2 = 0 + 2 \cdot 9.81 \cdot 2.5 = 49.05
\]
Now, take the square root to find \( v \):
\[
v = \sqrt{49.05} \approx 7.0 \, \text{m/s}
\]
Therefore, the speed of the object after it has traveled 2.5 meters downward is approximately **7.0 m/s**.