The Drawing Shows A Skateboarder Moving At 5.4 M/s

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Introductory Physics Homework Help

Question dealing with the conservation of mechanical energy.

Thread starter scroobnab
Start date Apr 9, 2011

Homework Statement

The drawing shows a skateboarder moving at 5.4 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is 0.40 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track.

Homework Equations

The Attempt at a Solution

I felt like my attempt was going well, but i can't seem to get the right answer. So i started out by realizing that Ef=Eo (i think), so KEf+PEf=KEo+PEo. im just confused as to how i should go about calculating thee final speed, which i think i need to be able to complete the equations for H.

Answers and Replies

also, sorry i dont have the picture attached, im new to the site and not even sure how to do that..

Can you put a number to the launch velocity (the skateboarder's speed as she leaves the end of the ramp)?

Can you put a number to the launch velocity (the skateboarder's speed as she leaves the end of the ramp)?

Thats one of my main questions... would that value be 5.4/cos48? so 8.07 m/s?

So with that in mind, when dealing with a problem involving KE, does the orientation of the velocity matter?

Thats one of my main questions... would that value be 5.4/cos48? so 8.07 m/s?
So with that in mind, when dealing with a problem involving KE, does the orientation of the velocity matter?

The orientation will become important in the next step. First you need to be able to find the correct launch velocity. This is determined by the kinetic energy that the skateboarder has when she reaches the top of the ramp. She begins on the horizontal with a given velocity, then rises up the ramp. While she goes up, she will be trading some kinetic energy for potential energy (due to gravity). Can you write a conservation of energy expression for this process?

Can you write a conservation of energy expression for this process?

so would the velocity at the end be:

1/2(m)(Vo)^2 + mgho = 1/2(m)(Vf)^2 + mghf

cancelling the m's, and isolating for Vf:

Vf = (2 (1/2(Vo)^2 - ghf))^1/2

if so, i calculated vf (at the end of the ramp) to be 5.02 m/s

You've got the right idea, but I think your algebra may have a bit of a problem.

1/2(m)(V_o)² + mgh_o = 1/2(m)(V_f)² + mgh_f

Cancel the m's and multiply through by 2, and take h_o = 0,

v_o ² = v_f ² + 2gh_f

v_f ² = v_o ² - 2gh_f

v_f = sqrt(v_o ² - 2gh_f)

Plugging in your given values:

v_f = 4.617 m/s (Keep an extra decimal place for intermediate results)

You've got the right idea, but I think your algebra may have a bit of a problem.

Oh okay, yeah stupid mistake. Alright now i'm getting a value of 4.6 m/s. So now i'm supposed to move on to the next section of her jump using this as the initial velocity?

Oh okay, yeah stupid mistake. Alright now i'm getting a value of 4.6 m/s. So now i'm supposed to move on to the next section of her jump using this as the initial velocity?

Right. Make a habit of hanging onto an extra decimal place or two for intermediate results in order to avoid rounding errors biting you in the end. Round results (answers) to the required precision.

So, the skateboarder leaves the ramp at the velocity that you've calculated, at the angle given for the ramp. This velocity can be broken into horizontal and vertical components, which can be treated independently. If you calculate the vertical component of the velocity, you should be able to use the appropriate equation of motion to determine the maximum height that she will reach.

okay bear with me here lol. so with that value i can use kinematics to solve for the rest right?

Vo = 4.617 m/s
ay = -9.8 m/s
Vf = 0 m/s
then solve for y?

okay bear with me here lol. so with that value i can use kinematics to solve for the rest right?
Vo = 4.617 m/s
ay = -9.8 m/s
Vf = 0 m/s
then solve for y?

Yes, that's the idea. Keep in mind that the Vo value is the composite speed; you need to break it down into its horizontal and vertical components, and use the vertical component to determine the maximum height.

Yes, that's the idea. Keep in mind that the Vo value is the composite speed; you need to break it down into its horizontal and vertical components, and use the vertical component to determine the maximum height.

Alright, i did this and got a value of 3.43 for Voy. after plugging the values into the equation my value for y was 0.6 m.

Thank you SO much for your help, this has been really helpful and has cleared up a lot of questions i had about Kinetic Energy and such. Cheers!