1-D Kinematics Equations involving time & distance

[razrsharp67]

Hey, I've been working on this problem for quite a while now, and I can't seem to come up with the solution. I would love some help in the right direction.

Homework Statement

Someone spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.61 s, and the top-to-bottom height of the window is 2.25 m. How high above the window top did the flowerpot go?

t_total= .61 s
Height = 2.25 m

Homework Equations

V_f = V_i + a_xt
X_f = X_i + 1/2(V_i + V_f)t

The Attempt at a Solution

To try and solve it, I figured that if the pot was in view for .61s total, it was .305s on each trip. So I figured its V_i = 2.25m/.305s = 7.377 m/s. So if its V_i=7.37, then it would take .7527 seconds to slow to 0m/s due to -9.8m/s² for gravity. After .7527 seconds it would fly to 2.77m, making it .53 meters above the window. Unfortunately, this is not the correct answer. Any suggestions?

 

[lewando]

You calculated an average velocity that is neither v_bot or v_top. Since you know a lot about x and t, why not use x = x₀ + v_x0t + a_xt²/2, (coordinate system: +x going up).

 

[razrsharp67]

Plugging my knowns into that equation still comes out to be the same thing though.

X = 0 + (7.377 * .7527) + (-9.8 * (.7527 ^ 2) * .5)
X = 0 + (5.55) + (-2.776)
X = 2.776
2.776 - 2.25 = .53 meters, which is still wrong.

 

[razrsharp67]

Is my problem coming from the fact that I can't actually split the .61s in half for each trip because they wouldn't be evenly distributed, making my initial velocity wrong? How do I fix that?

 

[gneill]

Is my problem coming from the fact that I can't actually split the .61s in half for each trip because they wouldn't be evenly distributed, making my initial velocity wrong? How do I fix that?

No, that's not your problem; The trajectory is symmetrical so it's split half and half as you've done.

Perhaps concentrate on finding the required velocities V_bot and V_top to begin with. You know that on the way up the pot will have some "initial" velocity at the bottom (V_bot) and must traverse the span of the window (known) in a given time (known). What equation governs the distance with respect to time?

 

[razrsharp67]

So, trying that:

2.25 = 0 + V_i * .305 + (.5 * -9.8 * (.305 ^ 2))
2.25 = .305V_i + (-.455)
2.70 = .305V_i
V_i = 8.87 m/s

t = -8.87 / -9.8 = .905 sec

X_f = 0 + 1/2 * (8.8 - 0) * .905 = 3.982 m

3.982m - 2.25m = 1.732 m

Which is still wrong. Ugh.

 

[gneill]

So, trying that:

2.25 = 0 + V_i * .305 + (.5 * -9.8 * (.305 ^ 2))
2.25 = .305V_i + (-.455)
2.70 = .305V_i
V_i = 8.87 m/s

t = -8.87 / -9.8 = .905 sec

Okay so far.

X_f = 0 + 1/2 * (8.8 - 0) * .905 = 3.982 m

Whoa! What equation is that? Why not use $d = v_o t - \frac{1}{2}g t^2$?

Alternatively, use energy conservation to find the height

 

[razrsharp67]

Whoa! What equation is that? Why not use $d = v_o t - \frac{1}{2}g t^2$?

So plugging into that one:

d = 8.8 * .905 - (.5 * -9.8 * (.905^2)) = 11.97 m
11.97 m - 2.25 m = 9.7m.

Which is still wrong. Thanks for the help Gneill, sorry I am so difficult.

 

[gneill]

The "-" sign in the equation was intended to make the acceleration negative. g is a positive constant (9.807 m/s²). The double negative has made the calculation invalid.

 

[razrsharp67]

So instead, its:

7.964 - .5 * 9.8 * (.905^2) = 3.950

3.950 - 2.25 = 1.7 meters above the window?

 

[gneill]

So instead, its:

7.964 - .5 * 9.8 * (.905^2) = 3.950

3.950 - 2.25 = 1.7 meters above the window?

You seem surprised. Why don't you verify by checking the velocity of a pot dropped from that height when it reaches the bottom sill of the window?

 

[razrsharp67]

1.7 still came back as incorrect :P

 

[gneill]

1.7 still came back as incorrect :P

You may need to retain a few more decimal places in your intermediate results and supply another significant figure in your answer.