Recent content by tchpowdog

Are you the guy that commented on my YouTube video? The "etc" part is solving for Y (which gives you the magnitude of the force needed). I will try to zip my Unity project up and send it to you somehow. It will be later today or tomorrow. I am having surgery in about 2 hours, so I will be in...

I agree. My x and y values are fixed. I started going down the wrong path with Vx being a given. All of these variables can't be givens if you want to clear the wall. You must solve for Vx and Vy (or Vi and angle).

I'm talking about deriving the equation of a parabola based on 3 points alone. This is before you introduce kinematics. Example: look at the image in my OP - if ##x_1=5m##, ##y_1=10m##, ##x_2=30m##, and ##y_2=5m##, you can see that the parabola faces up. The only purpose of the parabola is to...

Andy, please see my previous posts. I would also like to add more insight to this. The solution we've come to only works if your parabola faces down. There are cases where ##x_1## is really small (relative) and ##y_1 > y_2## resulting in a parabola facing up.. In that case, you must approach...

Is there a way to mark your comments as "answer"?

So I have figured all of this out. I had to change my approach a bit. In this scenario, ##V_x## and ##V_y## are both unknowns, so is angle ∅. With a given ##V_x##, you can either ensure that you clear the wall OR ensure that you hit your mark, but you can't do both. The parabola with 3...

So I was able to get my "a" and "b" coefficients to form a parabolic equation and I verified that they are correct values for my given points. From here, I know how to get the slope by taking the first derivative, but what do I do after that? equation: y = ax2 + bx + c y' = 2ax + b ← what...

I am solving the blue line based on the kinematic equations for time (time in the x direction, time in the y direction). When doing that, I have a given Vx, distance, and height - so solving for Vy is pretty simple. But when I introduce the wall (which can be anywhere between point A and B)...

<Rude comment deleted by mentor.> I don't care about the blue line, it's only there to show that the scenarios are different. If you don't want to help with the actual problem then stop replying, please.

... why?? Δy = ½gt2 + VyΔx/Vx Δy ⋅ 2Vx = ½gt2 ⋅ 2Vx + VyΔx/Vx ⋅ 2Vx 2VxΔy = Vxgt2 + 2VyΔx 2VxΔy - Vxgt2 = Vxgt2 + 2VyΔx - Vxgt2 2VxΔy - Vxgt2 = 2VyΔx Factor... Vx(2Δy - gt2) = 2VyΔx Vx = 2VyΔx / (2Δy - gt2)I'll go ahead and shame you now if you try to point out a typo in all that...

Δx = Vxt t = Δx/Vx Δy = ½gt2 + Vyt Δy = ½gt2 + VyΔx/Vx Solve for Vy. All other variables or known.

I have no desire to specify 3 points though. Point A - (0,y1) specified Point B - (x1, >y2) y component is unknown Point C - (x2,0) specified The y component of Point B is an unknown variable and I don't care what it is along as it's greater than y2 So how do I go about this? Any...

I think the problem with this is that x1 and x2 are not constant. They are variables that will have different inputs Everytime I run the equation in my program. Again, with given Vx, x1, x2, h1, and h2 - how do I solve for Vy so that the projectile clears the wall and lands at it's x2 position?

This is what is not true in my scenario. The projectile's height shouldn't be EQUAL to y2 at x1, the projectile's height just has to be GREATER than y2 at x1 so that it clears the wall AND hits it's x2 requirement. So how do I do that? (I just realized there was a sub script button lol)

Yes, I am calculating the blue trajectory based on functions of time and I believe this is the correct route to take, however, solving for Vy for X1 only does not consider the required X2 value. In other words, solving for Vy with required X1 and delta Y, the projectile would clear the wall but...