SimonJensen Posts: 1, Reputation: 1 New Member #1 Sep 11, 2017, 12:45 PM
Projectile Motion On An Inclined Plane
I was given the following information:
A ball is thrown with an initial velocity V0 atan angle of φ at an inclined planethat has an angle of θ with the horizontalline.

I really need and would appreciate some help to work of the following:
* The x-coordinate for the point where the ball touches the inclined plane (expressed by V0, g, θ and φ.

I have attached a link to an illustration I made of the problem.

https://ibb.co/hNKk6v

Thank you!
 ThomasCole Posts: 1, Reputation: 1 New Member #2 Sep 12, 2017, 06:07 AM
Want to know the answer too.
 ma0641 Posts: 15,681, Reputation: 1012 Uber Member #3 Sep 12, 2017, 10:21 AM
Must have the same homework.
 smoothy Posts: 25,495, Reputation: 2853 Uber Member #4 Sep 12, 2017, 08:57 PM
Both of you should consider vocational Tech curriculum if you can't do the homework in the one you are in.

The site rules REQUIRE you to show us your work and your answer, something neither of you have done.

Its a VIOLATION of site rules to hand out homework answers so you don't have to do your own work.
 MBarosM Posts: 5, Reputation: 1 New Member #5 Sep 13, 2017, 09:18 AM
I wasn't aware of the site's rules, so I apologize, as I was genuinely looking for help. Here is my own attempt to solve it, can anyone tell me if I am on the right track?:

My method was solving this equation:
0 = (tan(θ) - tan(φ)) - x( g / (2 * cos^2 (θ) * V0^2)

It has two solutions, either x = 0 or (tan(θ) - tan(φ)) - x( g / (2 * cos^2 (θ) * V0^2) = 0

Then I tried to solve the equation for x, and got the following as my result:
x = ((tan(θ)-tan(φ)) * (2 * V0^2 * Cos^2(theta)) / g when I try
 ebaines Posts: 12,130, Reputation: 1307 Expert #6 Sep 13, 2017, 10:46 AM
Originally Posted by MBarosM
My method was solving this equation:
0 = (tan(θ) - tan(φ)) - x( g / (2 * cos^2 (θ) * V0^2)
I get something a bit different:

$0 = \tan \phi - \tan \theta -x(\frac g 2 V_0^2\cos^2 \phi)$

It seems you have the definitions of phi and theta reversed. Also, you didn't specify this but I assume that the launch point is at the base of the incline, correct?
 MBarosM Posts: 5, Reputation: 1 New Member #7 Sep 13, 2017, 11:40 AM
Illustration of the situation: https://ibb.co/fSvwEF

Calculations:https://ibb.co/dtDBgv

If you are not able to read my handwriting, I did the following:

Y-direction: y = V0 * Sin(Theta) * t * 1/2 * g * t^2

X-direction: x = V0 * Cos(Theta) * t

I den solved for t in the equation for x and got:t = x / (V0 * Cos(theta)

I inserted the above value for t in the equation for y:

y = V0 * sin(theta) * (x / (V0 * cos(theta)) - g/2 *( x / (V0 * Cos^2(theta))

I substituted the left side of the equation (y) with tan(phi) * x and got:
tan(phi) * x = x * tan(theta) - g/2 * ( x / (V0 * Cos^2(theta))

I got the left side of the equation on the right side and got:
0 = x * (tan(theta) - tan(phi)) - x * ( g / (2 * V0^2 * Cos^2(theta))

After this, I solve the equation with respect to x (it is not on the attached picture), but I have done it:

0 = x * (tan(theta) - tan(phi) - x * ( g / (2 * V0^2 * Cos^2(theta))

x * ( g / (2 * V0^2 * Cos(theta))^2 = (tan(theta) - tan(phi))

x = ((tan(theta) - tan(phi)) * 2 * V0^2 * Cos^2(theta)) / g
 ebaines Posts: 12,130, Reputation: 1307 Expert #8 Sep 13, 2017, 01:48 PM
Originally Posted by MBarosM
Y-direction: y = V0 * Sin(Theta) * t * 1/2 * g * t^2

X-direction: x = V0 * Cos(Theta) * t
Two issues with this. First, the second half of that first equation should be "- 1/2 g t^2", not "* 1/2 g t ^2." This appears to be a simple typo, based on subsequent equations, so no harm done. The other issue is more critical: the equation of motion for the projectile is dependent on the angle of launch as measured from the horizontal. Where you use theta it should be theta + phi, because the sum of those two angles is the angle of the initial velocity V0 relative to horizontal. Please note that I made a similar mistake in my previous post - I hadn't had a chance to see the figure that was included with the original post, so assumed (incorrectly) that the angle phi was measured relative to horizontal. But now I see that phi is measured relative to the inclined plane, and hence the correct equation for Y-direction motion is:

y = V0 sin(theta + phi) t - (1/2) g t^2.

And for x-direction we have:

x = V0 cos(theta + phi) t.

We're interested in the point in time when the projectile hots the ramp, at which point y/x = tan(theta), so you can replace y in the first equation with x tan(theta). Can you take it from here?
 MBarosM Posts: 5, Reputation: 1 New Member #9 Sep 13, 2017, 02:05 PM
After I substitute tan(theta) * x for y in the equation, I have:

tan(theta) * x = tan(theta + phi) * x - g/2 * ( x^2 / (V0^2 * Cos^2 (theta + phi)

0 = tan(theta + phi) * x - tan(theta) * x - g/2 * ( x^2 / (V0^2 * Cos^2 (theta + phi)

0 = ( (-g * x^2) / (2 *cos^2(theta + phi) * V0^2)) + tan(theta + phi) * x^2 - tan(theta) * x

Am I on the right track now?
 ebaines Posts: 12,130, Reputation: 1307 Expert #10 Sep 14, 2017, 05:46 AM
Looking good. Now divide through by x and rearrange to get x by itself.

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