Question
ball dropped from 100m building speed calculation
Answers
Answered by
GPT-4o mini
To calculate the speed of an object dropped from a height (in this case, a ball dropped from a 100-meter building) just before it hits the ground, you can use the principles of physics, specifically the equations of motion under constant acceleration due to gravity.
Assuming there is no air resistance, the acceleration due to gravity (g) is approximately \(9.81 \, m/s^2\).
1. **Use the equation of motion:**
The final velocity (v) of an object under constant acceleration can be calculated using the following equation:
\[
v^2 = u^2 + 2as
\]
Where:
- \(v\) = final velocity (m/s)
- \(u\) = initial velocity (m/s) (which is 0 m/s if dropped)
- \(a\) = acceleration (m/s²) (which is \(9.81 \, m/s^2\))
- \(s\) = distance (m) (which is 100 m)
2. **Plug in the values:**
Since the ball is dropped, the initial velocity \(u\) is 0.
\[
v^2 = 0 + 2 \cdot 9.81 \cdot 100
\]
\[
v^2 = 1962
\]
\[
v = \sqrt{1962}
\]
\[
v \approx 44.37 \, m/s
\]
So, just before hitting the ground, the ball would be traveling at approximately **44.37 m/s**.
Assuming there is no air resistance, the acceleration due to gravity (g) is approximately \(9.81 \, m/s^2\).
1. **Use the equation of motion:**
The final velocity (v) of an object under constant acceleration can be calculated using the following equation:
\[
v^2 = u^2 + 2as
\]
Where:
- \(v\) = final velocity (m/s)
- \(u\) = initial velocity (m/s) (which is 0 m/s if dropped)
- \(a\) = acceleration (m/s²) (which is \(9.81 \, m/s^2\))
- \(s\) = distance (m) (which is 100 m)
2. **Plug in the values:**
Since the ball is dropped, the initial velocity \(u\) is 0.
\[
v^2 = 0 + 2 \cdot 9.81 \cdot 100
\]
\[
v^2 = 1962
\]
\[
v = \sqrt{1962}
\]
\[
v \approx 44.37 \, m/s
\]
So, just before hitting the ground, the ball would be traveling at approximately **44.37 m/s**.