Question
A cat chases a mouse across a tabletop. The mouse darts to the side at the last minute, causing the
cat to slide off the table. A diagram of the situation is shown below.
v0x=0.7 m/s
y=1.1 m
The cat is moving horizontally across the table 0.7 m/s. The table is 1.1 m above the ground.
Calculate the distance away from the table that the cat lands on the ground. Show all work.
You may use the Physics A equation sheet in the message board with the kinematic equations to solve
this problem. Assume no frictional forces. For simplicity, you may use the value of 10 m/s^2 as the
acceleration due to gravity
cat to slide off the table. A diagram of the situation is shown below.
v0x=0.7 m/s
y=1.1 m
The cat is moving horizontally across the table 0.7 m/s. The table is 1.1 m above the ground.
Calculate the distance away from the table that the cat lands on the ground. Show all work.
You may use the Physics A equation sheet in the message board with the kinematic equations to solve
this problem. Assume no frictional forces. For simplicity, you may use the value of 10 m/s^2 as the
acceleration due to gravity
Answers
GPT-4o mini
To solve this problem, we need to calculate how long it takes for the cat to fall to the ground and how far it travels horizontally during that time.
### Step 1: Calculate the time of flight
We can use the kinematic equation for vertical motion. The cat falls a vertical distance \(y = 1.1 \, \text{m}\) from the table to the ground. The equation we will use is:
\[
y = \frac{1}{2} g t^2
\]
where:
- \(y\) is the vertical distance (1.1 m),
- \(g\) is the acceleration due to gravity (approximately \(10 \, \text{m/s}^2\)),
- \(t\) is the time in seconds.
Rearranging the equation to solve for \(t\):
\[
1.1 = \frac{1}{2} (10) t^2
\]
\[
1.1 = 5 t^2
\]
\[
t^2 = \frac{1.1}{5}
\]
\[
t^2 = 0.22
\]
\[
t = \sqrt{0.22} \approx 0.469 \, \text{s}
\]
### Step 2: Calculate the horizontal distance traveled
The horizontal distance \(x\) that the cat travels while falling can be calculated using the horizontal velocity \(v_{0x} = 0.7 \, \text{m/s}\) and the time of flight \(t\) we just calculated:
\[
x = v_{0x} t
\]
Substituting the known values:
\[
x = (0.7 \, \text{m/s}) \times (0.469 \, \text{s})
\]
\[
x \approx 0.328 \, \text{m}
\]
### Conclusion
The cat lands approximately \(0.328 \, \text{m}\) away from the edge of the table.
### Step 1: Calculate the time of flight
We can use the kinematic equation for vertical motion. The cat falls a vertical distance \(y = 1.1 \, \text{m}\) from the table to the ground. The equation we will use is:
\[
y = \frac{1}{2} g t^2
\]
where:
- \(y\) is the vertical distance (1.1 m),
- \(g\) is the acceleration due to gravity (approximately \(10 \, \text{m/s}^2\)),
- \(t\) is the time in seconds.
Rearranging the equation to solve for \(t\):
\[
1.1 = \frac{1}{2} (10) t^2
\]
\[
1.1 = 5 t^2
\]
\[
t^2 = \frac{1.1}{5}
\]
\[
t^2 = 0.22
\]
\[
t = \sqrt{0.22} \approx 0.469 \, \text{s}
\]
### Step 2: Calculate the horizontal distance traveled
The horizontal distance \(x\) that the cat travels while falling can be calculated using the horizontal velocity \(v_{0x} = 0.7 \, \text{m/s}\) and the time of flight \(t\) we just calculated:
\[
x = v_{0x} t
\]
Substituting the known values:
\[
x = (0.7 \, \text{m/s}) \times (0.469 \, \text{s})
\]
\[
x \approx 0.328 \, \text{m}
\]
### Conclusion
The cat lands approximately \(0.328 \, \text{m}\) away from the edge of the table.