A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 s, and the top-to-bottom, height of the window is 2.00 m. How high above the window top does the flowerpot go?
The correct answer and explanation is :
To find how high above the window the flowerpot goes, we start by analyzing the motion of the flowerpot as it passes the window. It’s in free fall (under gravity), going up and then coming back down. Here’s the setup:
- The pot is visible for 0.50 s.
- The height of the window is 2.00 m.
- We’re asked to find how high above the window top the flowerpot travels before it turns around (its peak height).
Step-by-step:
Let’s assume:
- The top of the window is at position ( y = 0 ),
- The flowerpot enters the window going up, goes past the top, reaches its peak, and then comes down back past the window.
This is symmetric motion, so the pot spends half of the time going up past the window, and half coming down:
[
t_{\text{up}} = t_{\text{down}} = \frac{0.50}{2} = 0.25\, \text{s}
]
We now analyze one leg of the journey: from the top of the window up to the max height (0.25 s of motion). This is uniformly accelerated motion under gravity.
We use:
[
v_f = v_i – g t
]
At the peak, final velocity ( v_f = 0 ), so:
[
0 = v_i – g t \Rightarrow v_i = g t = (9.8)(0.25) = 2.45\, \text{m/s}
]
Now use:
[
h = v_i t – \frac{1}{2} g t^2
]
[
h = (2.45)(0.25) – \frac{1}{2}(9.8)(0.25)^2
]
[
h = 0.6125 – 0.30625 = 0.30625\, \text{m}
]
✅ Final Answer: The flowerpot rises approximately 0.31 meters above the top of the window.
Explanation (300 words):
This problem deals with an object in free fall observed within a window frame. The flowerpot is seen ascending and descending past the window, spending equal time going up and coming down. Since the flowerpot is only visible for 0.50 seconds, it spends 0.25 seconds on each leg of the journey.
Free fall motion follows the principles of uniformly accelerated motion under the influence of gravity. We assume the only force acting is gravity, and air resistance is negligible. By symmetry of the motion, the time to reach the peak (highest point) from the top of the window equals the time to fall back to that point.
We first calculate the initial velocity required for the pot to travel from the top of the window up to its maximum height in 0.25 seconds. Using the equation ( v = v_0 – g t ), and knowing that at the peak the velocity is zero, we find ( v_0 = g t = 2.45 \, \text{m/s} ). This is the speed with which the pot passed the top of the window upward.
Next, using this initial velocity, we compute the maximum height reached from the top of the window using the displacement equation ( h = v_0 t – \frac{1}{2} g t^2 ). Plugging in the values, we determine that the pot reaches a height of about 0.31 meters above the window.
This shows how kinematic equations, along with symmetry of projectile motion, allow us to analyze vertical motion even when we’re only given a brief moment of observation.