This is a problem from Halliday, Resnick, and Walker. My solution gets the correct answer, but it seems very extremely complicated and I think there is an easier way to do it.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

A 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?

2. Relevant equations

I'm not sure, maybe [tex]y(t)=\frac{1}{2}at^2+v_0t+y_0[/tex]

3. The attempt at a solution

I assumed that the time the flowerpot was in view going up and the time it was in view going down were equal, therefore they were both 0.25 seconds. How can I justify this?

Let y(t) be the height of the flowerpot at time t. Let y=0 m represent the top of the window, so y=-2 m would be the bottom of the window.

Let t=0 be the time the flowerpot first appears in the window, so y(0) = -2. The flowerpot reaches the top of the window at t=0.25 (by the assumption), so y(0.25) = 0.

Let t_{max}be the time when the flowerpot is at its highest point. Let y_{max}= y(t_{max})

Because the acceleration is constant, I know that the equation of the path of the flowerpot, y(t), is a parabola with the vertex at (t_{max}, y_{max}).

Therefore [tex]y(t) = c(t-t_{max})^2+y_{max}[/tex] for some constant c.

Multiply out y(t) to get: [tex]y(t) = c(t-t_{max})^2+y_{max} = c[t^2-2t_{max}t+t_{max}^2]+y_{max}=(c)t^2+(-2ct_{max})t+(ct_{max}^2+y_{max})[/tex]

Since the acceleration is due to gravity, the coefficient of t^{2}must be -4.9, so c=-4.9.

[tex]y(t) =(-4.9)t^2+(-2 \times -4.9 t_{max})t+(-4.9t_{max}^2+y_{max})=-4.9t^2+(9.8t_{max})t+(-4.9t_{max}^2+y_{max})[/tex]

We know the values of y(0) and y(0.25):

[tex]y(0) = -2 = -4.9(0)^2+(9.8t_{max})(0)+(-4.9t_{max}^2+y_{max})=-4.9t_{max}^2+y_{max}[/tex]

[tex]y(0.25) = 0 = -4.9(0.25)^2+(9.8t_{max})(0.25)+(-4.9t_{max}^2+y_{max})= -4.9(0.25)^2+(9.8t_{max})(0.25)+(-2)=-2.306+2.45t_{max}[/tex]

From [tex]-2.306+2.45t_{max}=0[/tex], we get [tex]t_{max}= 0.941[/tex] s

From [tex]-4.9t_{max}^2+y_{max} = -2[/tex], and substituting in t_{max}, we get [tex]y_{max}=2.34[/tex] meters above the window, which is the correct answer.

Wow, that took a long time to type and it looks like a mess. Does anyone know a cleaner solution? And how do I conclude that the time going up and down the window are equal? Thanks!

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# Homework Help: Flowerpot going up and down past a window

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