- #1

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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?

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- Thread starter NIZBIT
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- #1

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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?

- #2

Doc Al

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- #3

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- #4

Doc Al

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- #5

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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:

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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What's the 1.73 signify?NIZBIT said:

which is

sqrt(2*9.8*sin(30)*1.73=4.11 m/s^2

I still get the answer wrong.

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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- #10

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- #11

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- #12

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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.

- #15

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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].

(...Which perfectly fits the expression Doc Al gave: [tex]v^2 = v_0^2 + 2 a \Delta x[/tex].)

- #17

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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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