Recent content by v1ru5

I am still a little confused and we haven't gotten to the energy unit yet so I don't think it has anything to do with that. I was also thinking of using the equation Vf^2=Vi^2+2ad so I would get something like (1.4v)^2=(vsin40)^2+2*9.8*40. I am not sure if I am doing that correctly. I agree with...

Here are some projectile equations I have: 1) Vx = Vix (since no acceleration in x direction) 2) x = xi + vix*t (not useful since I do not know the displacement in the x direction) 3) Vy = Viy - g*t (cannot use this since I do not know t) 4) y = yi + Viy*t - 1/2*gt^2 (not useful since I do not...

Homework Statement A projectile is fired with an initial speed v at an angle 40 degrees above the horizontal from a height of 40m above the ground. The projectile strikes the ground with a speed of 1.4v. Find v. The Attempt at a Solution All I could find i Vinitialx=vcos40 and...

Thanks for your all your help Professor LCKurtz.

Oh okay, thanks for your help! See you around uOttawa :D

Oh okay so k~(x, y) = (kx - k + 1, ky + 2k -2) -1~(x,y) = -x + 1 + 1, -y -2 - 2 = -x + 2, -y - 4 So -(x, y) = (-x + 2, -y - 4) and that is correct since I got that as my answer when I did it the other way. So I can conclude that if I want to get the additive inverse of any vector, I...

anyone want to help me with part b please? I understand that it is asking whether k is a linear combination of g + h. So basically k = ag + bh of a and b are all real numbers. I just don't understand what number to use for a and b because none of them get be k :( Any help would be appreciated...

So if (x, y) is the vector in ##R^2## then to get its additive inverse I would multiply it by ##-1## which would give me (-x, -y). So if I do (x, y) ##+## (-x, -y) I should get the zero vector (1, -2). So (x, y) ##+## (-x, -y) (x - x - 1, y - y +2) = (1, -2) So x - x - 1 = 1 -1 ≠ 1 y...

so then since (a, b) is the inverse of (x, y) then (a, b) ⊕ (x, y) = 0 (a + x - 1, b + y +2) = (1, -2) a + x - 1 = 1 a = -x + 2 b + y + 2 = -2 b = -y - 4 so -v = (-x + 2, -y - 4) I can check this by doing v⊕(-v) = 0 so (x, y)⊕(a, b) = 0 (x + a - 1, y + b +2) = (1, -2) x...

so is that correct?

SO it would be: -(x, y) ⊕ (a, b) = 0 (-x, -y) ⊕ (a, b) = 0 (-x + a - 1, -y + b +2) = (1, -2) So -x + a - 1 = 1 a = x + 2 -y + b + 2 = -2 b = y + 4 So -v = (x + 2, y + 4)

Ya so (c, d) ⊕ (1, -2) = (c + 1 - 1, d + -2 + 2) = (c, d). I get it now :) thanks. Now for the second part. So 0 + v = v or v+(-v) = 0. I know that the zero vector (is 1, -2) so: (x, y) + -(x, y) = (1, -2) like that?

Oh so a = 1 and b = -2 so (1, -2) is the zero vector?

(x + a - 1, y + b +2) = (x, y) x + a - 1 = x a - 1 = 0 y + b + 2 = y b + 2 = 0 so (a-1, b+2) is the zero vector?

So then (x + a - 1, y + b +2) = (x, y)?