# Homework Help: Projectile Motion!

1. Oct 1, 2008

### mikee

1. The problem statement, all variables and given/known datasomeone standing on a down-tilting inclined plane making an angle of 30 degrees with the horizontal throws a ball with a horizontal velocity of 10m/s, how far down the plane will the ball strike ground

2. Relevant equations

3. The attempt at a solutionI was just wondering if i could just tilt the axis, so that my horizontal coordinate is now 10cos30degrees and my y coordinate is 10sin30degrees-gt?

then i plug can just use 10 for the Range equation and find when it hits the ground, so R= 10sin2(30degrees)/g? so i worked it out and came up with about 8.83 meters? is this way faulty?

2. Oct 1, 2008

### Wecht

Yes, absolutely you can. Designation of the axes and is arbitrary, meaning that you can choose the origin and the direction the axes point. Though it's probably a good idea to make sure they're perpendicular to each other.

3. Oct 1, 2008

### Wecht

BUT, you have to realize that the info bit 'horizontal' in the 'horizontal velocity' they give you is independent of your chosen coordinate system; meaning that if you 'tilt the axis', 'horizontal velocity' is no longer synonymous with x-velocity. What I mean is, you can change the coordinate system, but you can't change the magnitudes or the directions of the vectors you're given, so if you've put down the initial velocity as 10m/s x^ direction, and THEN you decide to change the direction of the axes (but keep the origin point), that initial velocity is no longer 10m/s x^direction; if you've changed the x-axis to be parallel with the incline, initial velocity will be also be changed (to 10m/s in a direction 30 degrees up from your new x-axis, in this case)

4. Oct 1, 2008

### Wecht

I'm also assuming that by 'x-coordinate is now 10cos30degrees' you mean x-velocity, and the same thing for 'y-coordinate'. Also realize that 'g' is no longer -9.81m/s^2 y-direction, if you've tilted the axes. Another point, don't you need to know what height above the incline the projectile starts from in order to get the final distance?

5. Oct 1, 2008

### mikee

well it does not give you the hieght so i think you can assume that it is 0, so when you tilt the axis the projectile is just going at a angle of 30 degrees relative to the new x axis so yea i understand when you tilt the axis you have to change the variable as needed, and also even though i tilted the axis the range equation is still the same right and gravity is also the same so i can still use sin2angle(Vo) / g? becuase g is always constant?

6. Oct 1, 2008

### mikee

Projectile Motion down an Inclined plane

1. The problem statement, all variables and given/known dataK just one quick question if an object is shot down an inclined plane at 10m/s horizontally at 30 degrees, if i put the x axis to be the inclined plane and the y axis and be perpendicular to the incline plane, would x(t) = 10cos30t and y= 10sin30t-1/2(gt^2) or would you have to add gravity in the x direction also because the coordinates were tilted?

2. Relevant equations

3. The attempt at a solution

7. Oct 3, 2008

### Wecht

The second way you put it is correct; as with the velocity, you must translate the acceleration into terms that fit your coordinate system, as -y-direction is no longer the only direction of acceleration.