Question
How far from the moving end of a 2.0 m board should a cup be placed to catch a ball rolling off at an angle of 30° when the support stick is suddenly removed?
Answers
GPT-4o mini
To solve the problem, we need to determine how far from the moving end of a 2.0 m board a cup should be placed to intercept a ball rolling off at an angle of 30° when the support stick is removed.
As soon as the support stick is removed, the board will begin to tilt and the ball will roll off the end of the board. Since we know the length of the board (2.0 m), we can calculate the time it will take for the ball to fall to the ground, and then we will use that information to find where the cup should be placed.
### Step 1: Calculate the height at which the ball leaves the board
Assuming the board is horizontal before it drops, when the stick is removed, the board will fall. Let's consider the vertical height at the pivot point when it's about to fall. This height can be approximated since it will fall a distance equal to the height from which the ball rolls off, which we assume is negligible until it fully pivots downwards.
However, the ball rolls off at an angle of \(30°\). To clarify, let's find the horizontal and vertical components of the velocity.
### Step 2: Break down initial velocity
If the ball falls off, we can assume it won't just drop but will roll off with the horizontal speed being the part parallel to the ground.
Let’s denote:
- \( v_{\text{initial}}\) = initial speed of the ball when it leaves the board at the tip.
### Step 3: Determine the time of flight
The time to fall can be calculated by considering the vertical drop when the ball falls off the end of the board.
\[
h = \frac{1}{2} g t^2 \implies t = \sqrt{ \frac{2h}{g} }
\]
Assuming the board is horizontal and the height \( h \) is \( L \sin(\theta) \), with \( L \) being the length of the board (2.0 m), and \( g \approx 9.8\ m/s^2\):
1. Find vertical drop height \( h \):
When the board starts to pivot, for small angles, the height \( h \) becomes negligible unless it's equal to the pivot height.
### Step 4: Calculate horizontal distance
Now, if the ball rolls off at an angle \(30°\), the horizontal distance covered while falling can be calculated by:
- \( d = v_{\text{horizontal}} t \)
Given that the horizontal component of velocity is,
\[
v_{\text{horizontal}} = v_{\text{initial}} \cdot \cos(30°)
\]
And vertical component dropping distance \( t =\sqrt{ \frac{2h}{g} }\)
### Conclusion:
Through calculation, for practical small board cases, we can say:
- The longer the ball rolls off the edge, the faster it moves horizontally.
- With the above, we get a newer distance for where to calculate the cup's placement.
If we take standard \( g \) and basic time estimates through height \( h \approx 0.1m\) results in placements concluding for horizontal distance.
### Final Calculation:
A more straightforward step would conclude:
The \( \text{total distance} = horizontal\ distance \to \text{ from pivot point} \approx 2 \cdot ( \text{time of fall} \cdot v \cdot cos(30°)) \).
This provides the overall distance for placement for the cup should be under factors as described along the forward path ensuring calculations rather discussed without integrating variables need refining based on final user provided speed inputs.
Thus concluding requiring \( d \) based on angular motion, optimally the distance would determine cup's placements possibly spanning \( \text{simple values as governed above} \).
As soon as the support stick is removed, the board will begin to tilt and the ball will roll off the end of the board. Since we know the length of the board (2.0 m), we can calculate the time it will take for the ball to fall to the ground, and then we will use that information to find where the cup should be placed.
### Step 1: Calculate the height at which the ball leaves the board
Assuming the board is horizontal before it drops, when the stick is removed, the board will fall. Let's consider the vertical height at the pivot point when it's about to fall. This height can be approximated since it will fall a distance equal to the height from which the ball rolls off, which we assume is negligible until it fully pivots downwards.
However, the ball rolls off at an angle of \(30°\). To clarify, let's find the horizontal and vertical components of the velocity.
### Step 2: Break down initial velocity
If the ball falls off, we can assume it won't just drop but will roll off with the horizontal speed being the part parallel to the ground.
Let’s denote:
- \( v_{\text{initial}}\) = initial speed of the ball when it leaves the board at the tip.
### Step 3: Determine the time of flight
The time to fall can be calculated by considering the vertical drop when the ball falls off the end of the board.
\[
h = \frac{1}{2} g t^2 \implies t = \sqrt{ \frac{2h}{g} }
\]
Assuming the board is horizontal and the height \( h \) is \( L \sin(\theta) \), with \( L \) being the length of the board (2.0 m), and \( g \approx 9.8\ m/s^2\):
1. Find vertical drop height \( h \):
When the board starts to pivot, for small angles, the height \( h \) becomes negligible unless it's equal to the pivot height.
### Step 4: Calculate horizontal distance
Now, if the ball rolls off at an angle \(30°\), the horizontal distance covered while falling can be calculated by:
- \( d = v_{\text{horizontal}} t \)
Given that the horizontal component of velocity is,
\[
v_{\text{horizontal}} = v_{\text{initial}} \cdot \cos(30°)
\]
And vertical component dropping distance \( t =\sqrt{ \frac{2h}{g} }\)
### Conclusion:
Through calculation, for practical small board cases, we can say:
- The longer the ball rolls off the edge, the faster it moves horizontally.
- With the above, we get a newer distance for where to calculate the cup's placement.
If we take standard \( g \) and basic time estimates through height \( h \approx 0.1m\) results in placements concluding for horizontal distance.
### Final Calculation:
A more straightforward step would conclude:
The \( \text{total distance} = horizontal\ distance \to \text{ from pivot point} \approx 2 \cdot ( \text{time of fall} \cdot v \cdot cos(30°)) \).
This provides the overall distance for placement for the cup should be under factors as described along the forward path ensuring calculations rather discussed without integrating variables need refining based on final user provided speed inputs.
Thus concluding requiring \( d \) based on angular motion, optimally the distance would determine cup's placements possibly spanning \( \text{simple values as governed above} \).