Final vertical velocity? (kinematic formulas)

  • Thread starter gibson101
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  • #1
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Here is the problem. It consists of a. b. and c. As you can see I found A and B, but C is giving me trouble.

My work is also attached. I am assuming that when the question says "with what speed do the supplies land in the latter case." that i am finding final velocity. and I have already found my initial which is 20.3. and this final velocity is going to be vertical, so the acceleration will be -9.8 m/s. And the distance (delta y is -235). This is a practice problem so the solution is given, to help me.
 

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  • #2
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Here is the work.
 

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  • #3
gneill
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Use energy conservation. The supplies are launched with some initial kinetic energy depending upon their speed at release. They will end up 235m below the launch height. What's the change in KE? How does this translate into final speed?
 
  • #4
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KE=1/2(mass)(v^2). It does not give me the mass, so how can I utilize this formula?
 
  • #5
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I've been on this problem for 2 hours, please some one help me. I just can't think anymore.
 
  • #6
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KE=1/2(mass)(v^2).

Correct.

It does not give me the mass, so how can I utilize this formula?

Don't worry about that for now. Keep it as an "m" in your equation. Maybe it will have disappeared by the time you solve for the variable you need.

Alternatively, you don't need to use energy at all. You know the initial velocity, the displacement, and the constant acceleration, and want to find the final velocity. Do you know a kinematic equation that relates these four quantities?
 
  • #7
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V^2=Vinitial^2+2(a)(deltay)
I've already tried this and it doesn't work.
v^2=(20.3)^2-4.9(-235)
v=39.5
 
  • #8
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acceleration is -9.8, delta y is -235, and Vinitial is 20.3
 
  • #9
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Your approach is definitely on the right track, but you have made two mistakes. First, you are substituting in 0.5*g rather than 2*g for some reason! Second, you have correctly found one component of the velocity, but this is not quite what the question is asking for. Can you see why not?
 
  • #10
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You're right! It should be -19.6. This is just my braining not performing right. Been at this for too long. It wants the speed. and my final velocity is 70.8. So i need the definition for speed?
 
  • #11
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Yep, although there's not much to it - speed is just the magnitude of velocity, so if the velocity were 70.8 m s^-1 downward the speed would just be 70.8 m s^-1. But is that the total velocity of the supplies?

(Just to clarify, your calculation leading to 70.8 m s^-1 is correct, but the answer doesn't mean quite what you think it does!)
 
  • #12
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Do i need to calculate for both x and y. cause i just calculate the final velocity for y. Do i need to set these up as vectors?
 
  • #13
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The question asks for the final velocity before it hits the ground, so the object won't just be going on the y direction, it will be going both x and y, correct?
 
  • #14
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You are absolutely correct that the final velocity is a vector with y component -70.8 m s^-1, so this certainly sounds like a sensible thing to do. Try it and see what happens!
 
  • #15
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Haha!!! r=squareroot of 45.8^2 + 70.8^2. R=84.3. YES! So happy! But now Im stuck on the same problem with just different values.
 

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  • #16
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Here is my work. For part B, i get -8.6 as my vertical initial velocity, which is incorrect.
 

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  • #17
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Well done - sterling work!

For the second one, read the problem again, carefully, and check that your answer means what you think it does. You have just shown that you understand exactly how to do these problems so I have absolutely no doubt you can figure out what's gone wrong here.
 
  • #18
gneill
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You should keep more decimal places in your intermediate results so that rounding errors don't spoil the accuracy of your final values.

Why did you enter 68.4 m/s as the required velocity when you calculated a value close to -8 m/s?
 
  • #19
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i have multiple attempts to submit an answer until its locked. I am employing the same methods that I used in the first problem, and i keep getting -8.6. I'm just hesitant to enter it in, because this is my last chance to get the correct answer before it locks up.
 
  • #20
gneill
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Using more decimal places for g and Δt I calculate, using the same method as you, a slightly smaller value for the velocity. It might be significant if the "marker" is picky.
 
  • #21
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-8.64 has to be incorrect for part B, because when i used -8.6 for part c i get 106.1 which is wrong.
 

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  • #22
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no its not picky, sig figs do not matter, and ignore the 68.4, it was another answer i somehow got, but its wrong.
 
  • #23
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gneill, i'm assuming you got -8.37. That is the correct answer.
 
  • #24
gneill
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gneill, i'm assuming you got -8.37. That is the correct answer.

Hi gibson, my value was -8.39 m/s, but I keep lots of decimal places in all intermediate results and use 9.80665 m/s2 for g (I'm using a math spreadsheet, so this is a no-brainer!). Your value looks fine.

When I have to work "by hand", whenever I see formulas where values are being squared I make a point of keeping extra decimal places, because squaring small errors makes for much bigger errors!
 

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