Projectile motion skateboarder help

In summary, the skateboarder starts up a 1.0-m-high, 30deg ramp at a speed of 7.0 m/s. The skateboard wheels roll without friction.
  • #1
NIZBIT
69
0
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?
 
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  • #2
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
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
If you want to find the speed at the top, use conservation of energy. Or find the acceleration up the ramp.
 
  • #5
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
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
I'm guessing -9.8? But I don't know the time so how can I use that?
 
  • #8
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
Well honestly, I 've done this problem so much I really don't know anymore. I don't know if accelreation is 9.8 or -gsin(theta). The displacement I'm assuming is the hypotenuse which is 2.
 
  • #10
You're right, take g sin theta for the acc (remember it's negative) and the displacement along the ramp is 2.
 
  • #11
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
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
I did sqrt((7^2)+(2*-9.8sin(30)*2)).
 
  • #14
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
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
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
And which is why I said I didn't agree with 6.26 m/s
 
  • #18
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:
 

Related to Projectile motion skateboarder help

1. What is projectile motion?

Projectile motion is the motion of an object through the air, under the influence of gravity. It is a combination of horizontal and vertical motion, and is affected by the initial velocity, angle of launch, and the force of gravity.

2. How does projectile motion apply to skateboarding?

In skateboarding, projectile motion is seen when a skateboarder launches off a ramp or performs a trick such as an ollie. The skateboarder's body and the skateboard follow a curved path through the air, similar to the motion of a projectile.

3. What factors affect the distance and height of a skateboarder's jump?

The distance and height of a skateboarder's jump are affected by the initial velocity, angle of launch, and the force of gravity. A higher initial velocity and a greater angle of launch will result in a longer and higher jump, while the force of gravity will cause the skateboarder to eventually fall back to the ground.

4. How can friction impact a skateboarder's motion?

Friction can impact a skateboarder's motion in several ways. When rolling on a flat surface, friction between the skateboard wheels and the ground can slow down the skateboarder's motion. However, in some cases, friction can also be used to the skateboarder's advantage, such as when performing a slide or stopping the skateboard.

5. How can understanding projectile motion help improve skateboarding performance?

Understanding projectile motion can help improve skateboarding performance by allowing skateboarders to predict and control their motion more accurately. By adjusting the initial velocity and angle of launch, skateboarders can achieve greater distances and heights in their jumps. Additionally, understanding the effects of friction can help skateboarders make more precise movements and maintain better control of their board.

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