How Do You Calculate the Optimal Launch Angle for a Marble on a Sloped Track?

[lax1113]

Hey guys,
I am wondering if anyone could point me in the right direction as to how one would solve this problem, not just to get an answer, but to be able to get an answer for different distances. I have gotten the answer by guess and check, but that is pretty much useless for understanding the concept.

It was essentially a lab we are doing currently, where we have marbles on a track. The object is to get the marble to leave the track and land on a certain point, X distance away from the table. (the track is on a table, the table is roughly a meter high... so it is one side of the track elevated by books, then it slopes down and goes even with the table until it is to the end of the table where it woudl leave, and go towards the point on the floor). So to do this, we can calculate the velocity of the marble at any point of the track using a piece of equiptment.

With that backround info, the idea of the problem is to prop the one end of the track (the end the marble is leaving at) to the right angle so that the marble will have the perfect velocity in the x and y vector to land at the point.

So far I have tried to find a way to have the angle solved in terms of x and y, so that when i measure the y displacement, or the height of the table and the track, and the x displacement, the distance from the point on the floor to the end of the track, I could get an angle that would make the marble land in that point.

As of now I have only been able to guess with angles and then plugging them in get a displacement for x, then if its more or less than i need, adjust the angle accordingly. A few times i have been working through equations and ill think i have it, only to go brain dead at the last step I woudl really like most of all to just get a little nudge because i am very close to solving it.

For the equaitons, the hypotenuse is measured as 27.5cm, this is the length of the track before moving it in any angle, so it remains 27.5cm after i alter the angle.

By the way, although I am pretty sure that an equation can be made so that it solves for the angle without the velocity, the velocity of the marble before altering the angle is 1.6m/s. (that was measured with equiptment before the end of the track was altered)

equations-
y=VoT+.5GT²
x=Vx(t)
Vy=sin(0)*27.5cm
Vx=cos(0)*27.5cm

http://img442.imageshack.us/my.php?image=physicsld1.png

http://img442.imageshack.us/img442/539/physicsld1.th.png

Kinda small but if you click on it it should be bigger. Hopefully will give a better picture of what i was trying to say in words.

 

[madmike159]

You need the speed as it leaves the ramp, its max hight above the tabel and its acceleration (its start speed is 0). From that you should be able to use Newton's laws of motion (ignoring ari resistance) and some trig to work out the best angle.

 

[baba89]

Hi there,

It sounds like you are working on a projectile motion problem, specifically with a marble on a track. To solve this problem, you will need to use equations of motion for both the x and y components of the marble's motion.

First, let's define the variables we will be using:

- x = horizontal distance traveled by the marble (in meters)
- y = vertical distance traveled by the marble (in meters)
- g = acceleration due to gravity (9.8 m/s^2)
- t = time (in seconds)
- Vx = initial velocity of the marble in the x direction (m/s)
- Vy = initial velocity of the marble in the y direction (m/s)
- θ = angle of the track (in degrees)

To start, we can use the equation y = Vyt + 1/2gt^2 to solve for the time it takes for the marble to reach the end of the track. Since we know the initial velocity in the y direction (Vy) and the vertical displacement (y), we can rearrange the equation to solve for t:

t = √(2y/g - 2Vy/g)

Next, we can use the equation x = Vxt to solve for the time it takes for the marble to reach the end of the track in the x direction. Since we know the initial velocity in the x direction (Vx) and the horizontal displacement (x), we can rearrange the equation to solve for t:

t = x/Vx

Now, we can set these two equations equal to each other and solve for the angle (θ):

√(2y/g - 2Vy/g) = x/Vx

θ = arctan((xVx)/(2y - 2Vy))

You can now use this equation to calculate the angle needed for the marble to land at a specific point (x,y) on the floor. Keep in mind that this equation assumes a flat track, so if the track is elevated at any point, the calculations will be slightly different.

I hope this helps, and good luck with your lab! Remember to always double check your calculations and make sure your units are consistent. Keep practicing and you will get the hang of it.