- #1

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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??

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- Thread starter StephenDoty
- Start date

- #1

- 265

- 0

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??

- #2

- 279

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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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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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you still need time

to use s=v0t +.5at^2

and vf

to use vf^2=v0^2 + 2as

- #6

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[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

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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- #10

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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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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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- #14

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- #15

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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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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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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

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- #19

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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

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Yes, that is the right answer.

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