Question
A 50kg object is dropped from a 60m high building. If the object leaves a 10cm impression in the ground, what was the speed of the object when it hit the ground and how long did it take to fall?
Answers
GPT-4o mini
To solve the problem, we will calculate the speed of the object just before it hits the ground and the time it takes to fall from the height of the building.
### Step 1: Calculate the speed just before impact.
We can use the following kinematic equation that relates distance fallen \( d \), initial velocity \( v_0 \), acceleration \( a \), and final velocity \( v \):
\[
v^2 = v_0^2 + 2ad
\]
Since the object is dropped (not thrown), the initial velocity \( v_0 = 0 \). The acceleration \( a \) is due to gravity, approximately \( 9.81 \, \text{m/s}^2 \), and the distance \( d = 60 \, \text{m} \).
Substituting the known values into the equation:
\[
v^2 = 0 + 2 \cdot 9.81 \cdot 60
\]
\[
v^2 = 1177.2
\]
\[
v = \sqrt{1177.2} \approx 34.3 \, \text{m/s}
\]
### Step 2: Calculate the time it takes to fall.
We can use another kinematic equation to find the time \( t \):
\[
d = v_0 t + \frac{1}{2} a t^2
\]
Again, since \( v_0 = 0 \):
\[
60 = 0 + \frac{1}{2} \cdot 9.81 \cdot t^2
\]
Rearranging it gives:
\[
60 = 4.905t^2
\]
\[
t^2 = \frac{60}{4.905} \approx 12.22
\]
\[
t \approx \sqrt{12.22} \approx 3.49 \, \text{s}
\]
### Summary of Results
- The speed of the object just before it hits the ground is approximately **34.3 m/s**.
- The time it takes to fall from the building to the ground is approximately **3.49 seconds**.
Please let me know if you need any further assistance!
### Step 1: Calculate the speed just before impact.
We can use the following kinematic equation that relates distance fallen \( d \), initial velocity \( v_0 \), acceleration \( a \), and final velocity \( v \):
\[
v^2 = v_0^2 + 2ad
\]
Since the object is dropped (not thrown), the initial velocity \( v_0 = 0 \). The acceleration \( a \) is due to gravity, approximately \( 9.81 \, \text{m/s}^2 \), and the distance \( d = 60 \, \text{m} \).
Substituting the known values into the equation:
\[
v^2 = 0 + 2 \cdot 9.81 \cdot 60
\]
\[
v^2 = 1177.2
\]
\[
v = \sqrt{1177.2} \approx 34.3 \, \text{m/s}
\]
### Step 2: Calculate the time it takes to fall.
We can use another kinematic equation to find the time \( t \):
\[
d = v_0 t + \frac{1}{2} a t^2
\]
Again, since \( v_0 = 0 \):
\[
60 = 0 + \frac{1}{2} \cdot 9.81 \cdot t^2
\]
Rearranging it gives:
\[
60 = 4.905t^2
\]
\[
t^2 = \frac{60}{4.905} \approx 12.22
\]
\[
t \approx \sqrt{12.22} \approx 3.49 \, \text{s}
\]
### Summary of Results
- The speed of the object just before it hits the ground is approximately **34.3 m/s**.
- The time it takes to fall from the building to the ground is approximately **3.49 seconds**.
Please let me know if you need any further assistance!