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2-d motion of skateboarder

  • #1
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A skateboarder starts up a 1.0-m-high, 30 ramp at a speed of 7.0 m/s. The skateboard wheels roll without friction. How far from the end of the ramp will the skateboarder touch down?

I used:
Vy=7sin30=3.5
Vx=7cos30=6.1
vf^2=(3.5^2) - 19.6(-1)
vf=5.64

what do I do now??
 

Answers and Replies

  • #2
279
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You've caulculated his horizontal and vertical velocities the instant he hits the ramp. You need to calculate his horizontal and vertical velocities the instant he leaves the ramp.

To do this you'll need to determine his kinetic energy at the top of the ramp, which you can find from his initial kinetic energy and his gravitational potential energy at the top of the ramp.

Although you don't have the mass of the skateboarder, you don't need it since the mass will cancel out.
 
  • #3
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I havent learned KE and PE
 
  • #4
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OK then, another way you could do it is to find the length of the ramp, then calculate how much his speed slows while travelling along it.

His speed is initially 7ms, he travels along the ramp with a force of mgSin30 acting to slow him down. You can calculate his deceleration with F = ma, notice the mass cancels. Then with his initial velocity, his deceleration and the length of the ramp, you can determine his final velocity at the top of the ramp.

Then you can resolve that into vertical and horizontal velocity and find out how far he travels horizontally.
 
  • #5
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but to find the length with a=gsin30 and v0= 7m/s
you still need time
to use s=v0t +.5at^2
and vf
to use vf^2=v0^2 + 2as
 
  • #6
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As I said, you have his initial velocity, his deceleration and the length of the ramp. Therefore you can use:
[tex]v^{2} = u^{2} + 2as[/tex]

Where v is final velocity (what you're trying to find), u is initial velocity, a is his acceleration (which in your case will be negative) and s is the displacement. Time doesn't factor into this equation.
 
  • #7
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but how do u find the displacement without time
 
  • #8
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You're actually given the displacement in the problem(or the info necessary to find it, I'm thinking you meant to say it's 1 m high with a 30 degree angle?), he's traveling up a ramp, you can find how long the ramp is

You could actually use the displacement, acceleration, and initial velocity to find how long it takes him to travel up the ramp but it's not asked for and it's not needed

Remember, the goal for this part is to just find his velocity when he leaves off the top of the ramp, then you can treat it like a typical 2-d projectile problem
 
Last edited:
  • #9
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Im missin something. I just dont understand how knowing the angle and the height of the ramp gives me how long the ramp is which I can use to find final velocity off the ramp or the initial velocity of the second part? You could have a 5 m ramp that is 1 meter high and 30 degrees or you could have a 2m ramp that is 1 meter high and 30 degrees.
 
  • #10
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No you couldn't, because the ramp forms a triangle with the ground. In fact it's a right angled triangle which makes things even easier.

You have one angle and the side opposite that angle, and you want to find the hypotenuse.
 
  • #11
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I just dont understand how knowing the angle and the height of the ramp gives me how long the ramp is
well trig, it's a right triangle. You kinda lose me after that, draw it. If the ramp is 1 m high and makes an angle of 30 degrees with the horizontal, there is one answer in all of heaven and earth for the length of that ramp, and I think you're thinking it's ambiguous for some reason. In that sentence when I said "length of the ramp" I think you know what I meant but to be clear, the actual RAMP ramp length, the hypotenuse of the triangle

Once you discover the length of the ramp(ie the actual distance the skateboard..ist travels)you know the acceleration acting to slow him down that the other poster mentioned and you know his initial velocity upon contact with the ramp.

Using a like the previous poster mentioned, you know the length of the ramp, a, and initial velocity, so you can find his velocity at the point where he launches off the ramp(let's call it V)

Then the problem becomes "A projectile is launched 1m above the ground at an angle of 30 degrees with an initial velocity of V"
 
  • #12
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duh...didnt even think about that.
so the ramp is 2m
thus giving the velocity going off the ramp as 8.28m/s
using
vy=4.14m/s
vx=7.17m/s
vf^2=(4.14)^2 + (-9.8)(-1), vf= 5.19m/s downward=-5.19m/s
and thus -5.19=4.14 - 9.8t making t= .95s

Thus d=vt = 7.17 * .95 = 6.8m

Is this right??
 
  • #13
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No, as I said, he will decelerate as he goes up the ramp, therefore his speed at the top MUST be lower than his speed at the bottom.
 
  • #14
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er, remember that a should be -g*sin(30), I'm betting you forgot the negative. He hits the ramp at 7 m/s and then leaves the ramp going faster?
 
  • #15
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the ramp is 2m
thus giving the velocity going off the ramp as 5.42m/s
by vf^2= 49 -4(9.8)sin30=49-2(9.8)=29.4=5.42
using
vy=2.71m/s
vx=4.69m/s
vf^2=(2.71)^2 + (-9.8)(-1), vf= 4.14m/s downward=-4.14m/s
and thus -4.14=2.71 - 9.8t making t= .699s

Thus d=vt = 4.69 * .699 = 3.28m

I put this into mastering physics and it was wrong? What did I do wrong?
 
  • #16
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Closer this time, but still not quite there. Remember the skater goes up at 2.71 m/s until he comes to a rest, then falls to the ground from there. You have to take into account the time it takes him to reach his maximum height, then calculate the maximum height so you can also calculate the time he takes to fall it. Then add these times together to find his total air time.

Edit: Also, the formula is:
[tex]v^{2} = u^{2} + 2as[/tex]

You seem to have forgotten about the 2.
 
  • #17
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ok so...
I used vf=v0+ at
0=2.71-9.8t
t=.277s

then

to find the time from max height
vf^2=2as=-19.6 *-1 = 19.6=4.43m/s
then
4.43=v0+at
4.43=at t= .45s

total time= .277 + .45 = .73s
.73s * 4.69m/s= 3.42m

is this right?
 
  • #18
279
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No, because you never found the maximum height, you used 1m again. As I said the skater is still going up when he leaves the ramp, so his max height is higher than 1m.
 
  • #19
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ok so...
I used vf=v0+ at
0=2.71-9.8t
t=.277s

then
max height: 0=(2.71)^2 -19.6 s, 7.34= 19.6s, s=.37m + 1m=1.37m
to find the time from max height
vf^2=2as=-19.6 *-1.37 = 26.85= 5.18m/s
then
5.18=v0+at
5.18=at t= .52s

total time= .277 + .52 = .81s
.81s * 4.69m/s= 3.8m

is this right now?
 
  • #20
279
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Yes, that is the right answer.
 

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