Projectile motion skateboarder help

  • Thread starter NIZBIT
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  • #1
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This problem is not fun:
Code:
A skateboarder starts up a 1.0-m-high, 
30deg ramp at a speed of 7.0 m/s. 
The skateboard wheels roll without friction.

I guees the first thing I'll ask is the 7 m/s at 30 deg the launch velocity and angle?
 

Answers and Replies

  • #2
Doc Al
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No, 7 m/s is the speed at the bottom of the ramp. You'll need to figure out the speed when it takes off at the top.
 
  • #3
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ok, I don't know how to resolve this. I understand that the skateboarder is going to be slower at the top and gravity is the only force stopping motion.
 
  • #4
Doc Al
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If you want to find the speed at the top, use conservation of energy. Or find the acceleration up the ramp.
 
  • #5
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I have not learned the conservation of energy yet. I just used the formula sqrt(2*g*sin(theta)*delta(x)
which is
sqrt(2*9.8*sin(30)*1.73=4.11 m/s^2

I still get the answer wrong.
 
  • #6
radou
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NIZBIT said:
I have not learned the conservation of energy yet. I just used the formula sqrt(2*g*sin(theta)*delta(x)
which is
sqrt(2*9.8*sin(30)*1.73=4.11 m/s^2

I still get the answer wrong.

Use v = Vo + at, where Vo is your initial speed, and a the acceleration. Now, what does this acceleration equal while the skater is climbing up the ramp?
 
  • #7
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I'm guessing -9.8? But I dont know the time so how can I use that?
 
  • #8
Doc Al
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NIZBIT said:
I have not learned the conservation of energy yet. I just used the formula sqrt(2*g*sin(theta)*delta(x)
which is
sqrt(2*9.8*sin(30)*1.73=4.11 m/s^2

I still get the answer wrong.
What's the 1.73 signify?

What you are using here--and which is perfectly OK if done right--is the formula:
[tex]v^2 = v_0^2 + 2 a \Delta x[/tex]

To use it properly, answer these questions:
What's the acceleration along the ramp?
What's the displacement along the ramp?
What's the initial velocity?
 
  • #9
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Well honestly, I 've done this problem so much I really don't know anymore. I dont know if accelreation is 9.8 or -gsin(theta). The displacement I'm assuming is the hypotenuse which is 2.
 
  • #10
rsk
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You're right, take g sin theta for the acc (remember it's negative) and the displacement along the ramp is 2.
 
  • #11
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Ok so I got 6.26 m/s. Now the problem becomes projectile motion I hope. Now I just need to find the range.
 
  • #12
rsk
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That's not what I get.....can you show me your calculation.

As far as I can see, both your distance up the slope and the acceleration with have sin30 in them, so the sin30s will cancel.
 
  • #13
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I did sqrt((7^2)+(2*-9.8sin(30)*2)).
 
  • #14
radou
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NIZBIT said:
I did sqrt((7^2)+(2*-9.8sin(30)*2)).

Seems correct, as far as I'm concerned.. I did another 'method' too, to check. :smile:
 
  • #15
rsk
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NIZBIT said:
I did sqrt((7^2)+(2*-9.8sin(30)*2)).

That gives me 49 - 19.6 which gives me 29.4. The sq rt of that is not 6.26
 
  • #16
radou
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rsk said:
That gives me 49 - 19.6 which gives me 29.4. The sq rt of that is not 6.26

[tex]\sqrt{7^2 - 2\cdot9.81\cdot\sin(30)\cdot2} \approx 5.42[/tex]. :rolleyes:

(...Which perfectly fits the expression Doc Al gave: [tex]v^2 = v_0^2 + 2 a \Delta x[/tex].)
 
  • #17
rsk
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And which is why I said I didn't agree with 6.26 m/s
 
  • #18
radou
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rsk said:
And which is why I said I didn't agree with 6.26 m/s

Ooops, sorry, I seem to have misread you. :shy:
 

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