Why Does My Projectile Land Incorrectly After Launching Off a Ramp?

In summary, The problem involves a skateboarder starting on a 1.0-m-high, 30 degree ramp at a speed of 7.0 m/s. The skateboard wheels have no friction, but the skateboarder is having trouble accurately calculating the distance they will travel before touching down. They have attempted various methods, including using the y-component of motion and the quadratic formula, but have not been successful. Other users suggest using a new coordinate system, separating the weight of the skater into y and x components, and using conservation of energy to calculate the final velocity. Eventually, the user is able to calculate the correct answer, but expresses frustration with the challenging nature of the problem.
  • #1
FancyNut
113
0
This problem looks so simple but every answer I get is wrong... :eek:

A skateboarder starts up a 1.0-m-high, 30 degree ramp at a speed of 7.0 m/s. The skateboard wheels roll without friction.

How far from the end of the ramp does the skateboarder touch down?



EDIT : I realized several mistakes.. this is how I did it the last time:

I just tried it again... this time I calculated how long the ramp is which is 2 (since hight is 1 and sin 30 = 1/2). Now using this distance I calculated the final velocity the dude leaves the ramp with... then using that velocity I took the y-component of it and calcuated the time it takes to hit ground... used that time to calculate final position... and it was wrong.


I used the y-component of motion (free fall) to calculate time..

[tex]0 = 1 + v_0 sin (30) t- \frac {g}{2} t^2[/tex]


then I plug into this:

[tex]x_f = v_0 cos (30) t[/tex]


I have no idea what I'm doing wrong here... Ancy help would be much appreciated.
 
Last edited:
Physics news on Phys.org
  • #2
i think you chose the wrong y and x axises. i suggest you to choose the x-axis directed along the ramp surface line. then you will have to separate the mg force to y and x components. it will help me a lot, if you also type the answer which is written in your workbook. :smile:
 
  • #3
I'll try that new coordinate system...

I don't have the final answer. I'm using www.masteringphysics.com and the only way to get it is to request it which will end the problem for me-- 10 points off. :(
 
  • #4
Ok I guess I should've calculated the final velocity the skater takes off..

acceleration is - 9.8 sin 30 and after doing everything again it's still... wrong.
 
  • #5
ok, then we go to plan b. can you give me more details, such as the weight of the skater, how long is the ramp and etc?
 
  • #6
There is no weight given.

ugh, my head hurts. :(

I just tried it again... this time I calculated how long the ramp is which is 2 (since hight is 1 and sin 30 = 1/2). Now using this distance I calculated the final velocity the dude leaves the ramp with... then using that velocity I took the y-component of it and calcuated the time it takes to hit ground... used that time to calculate final position... and it was wrong. :frown:
 
  • #7
I think I have it right this time except one thing... this second degree equation. I used the quadratic formual... and after applying it and using the new t value I got the answer is STILL wrong. Why does this have to happen. :cry:

[tex] 0 = 1 + 14.7 t - \frac {g}{2} t^2[/tex]

[tex] 4.9 t^2 - 14.7 t - 1 = 0 [/tex]
 
Last edited:
  • #8
anybody... ?
 
  • #9
Where did u get 14.7 from?

Try using v^2=vi^2+2ad

To get the speed at the top of the ramp.
 
  • #10
I recommend you use conservation of energy.

We have MVinitial^2/2=MVfinal^2/2+Mgh
V^2=Vfinal^2+2gh

where Vfinal - speed at top of ramp, g - acceleration of gravity - h height of ramp. once you have vfinal, you can calculate its y component by trig and see how long it'll take for the skater to go up and down to the Earth again (under acceleration -g), and then use that time to calculate the horizontal distance from the horizontal component of Vfinal
 
  • #11
whew. I finally got the right answer. Thanks guys!

Is it me or is masteringphysics.com the scariest educational tool in existence? Typing in my number and pressing the 'submit' tab while dreading the 'try again' message is a horrible, horrible experience. :(
 

Related to Why Does My Projectile Land Incorrectly After Launching Off a Ramp?

What is kinematics in two dimensions?

Kinematics in two dimensions is the branch of physics that deals with the motion of objects in a two-dimensional space. It involves studying the position, velocity, and acceleration of an object as it moves along a two-dimensional path.

What are the key concepts in kinematics in two dimensions?

The key concepts in kinematics in two dimensions include displacement, velocity, acceleration, and projectile motion. Displacement is the change in position of an object, velocity is the rate of change of displacement, and acceleration is the rate of change of velocity. Projectile motion refers to the motion of an object that is launched into the air and moves along a curved path under the influence of gravity.

How do you calculate displacement in two dimensions?

To calculate displacement in two dimensions, you can use the Pythagorean theorem. This involves finding the square root of the sum of the squares of the vertical and horizontal displacements. Another method is to use vector addition, where you add the individual displacements in the x and y directions to get the resultant displacement.

What is the difference between speed and velocity in two dimensions?

Speed is a scalar quantity that refers to how fast an object is moving, while velocity is a vector quantity that includes both the speed and direction of an object's motion. In two dimensions, velocity also includes the direction of the object's motion in the x and y directions.

How is acceleration calculated in two dimensions?

To calculate acceleration in two dimensions, you can use the formula a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time interval. This formula can be applied separately for the x and y directions, as acceleration can vary in different directions in two dimensions.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
2
Replies
56
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
987
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
367
  • Introductory Physics Homework Help
Replies
25
Views
626
Back
Top