So if I were to solve it in reverse, it would be PE = PE + KE.
All variables are known at that point, except velocity. Makes sense! It's also extremely simple...
It also means that I have to re-evaluate this problem tomorrow.
Took a break, but I've solved for d:
So if I know the velocity in the Y, it turns into a trig problem. So I can find the x using tanθ=6.32/a
So the velocity in the x axis is 3.65m/s.
Using that, I can easily solve for D.
3.65*4 = 14.6m
Now I just need to solve for θ2:
All I need...
Okay... While I don't know the initial velocity, I know that it's going to hit the top of the building with a certain velocity. The velocity at 20m off the ground will be the same for both sides of the curve.
So I can use KEi=PE
V (at 20m) = 6.32m/s
I'm going to be a tad upset if I just figured this out:
Using kinematics from the top of the parabola:
for the Y
Vf=Vi + at
Vf= 0 + 10*4
Plug into equation from my first post:
mgH= 1/2*mv^2 + mgh
H= (1/2*v^2 + gh)/g
No, this isn't right either! It's not 4 seconds...
Well, I took my Physics final today, and I passed it but I didn't do quite as well as I'd hoped. Anyway, there was one problem I spent tons of time trying to figure out. I've tried giving it a shot again at home, but it's still eluding me. Some guidance would be great! Sorry if this is...
Note: Solving for part D and E
Unsure. Language in the problem seems ambiguous.
Perhaps (KE + PE)b = (KE + PE)a
The Attempt at a Solution
So I think it's done this way.
From A to B and...
Hello, new to these forums! I'm working on my UPI homework, and I just want to verify if I'm doing something correctly. The problem:
The position vector r of a particle moving in the x - y plane given by r = (2t^3)i + (6-7t^4)j, where r is in meters and t is in seconds.