So the equations for the X and Y velocity given $\theta$ and $V_0$ are $V_x = V_0\cos\theta$, and $V_y = V_0\sin\theta$. When I test this with something like 1 m/s and and angle of $45^{\circ}$, I get a really weird result. Logically, the $x$ and $y$ velocities should add to $V_0$. Unless I have done something wrong, they do not add to $V_0$. Can anyone explain this? How these equations can be accurate if $V_0\neq V_x + V_y$?

The only explanation I can think of is that the equations are incorrect in the first place. Does anyone have a solution to this problem?

  • 1
    $\begingroup$ In a triangle, the largest side must be smaller than the other two sides combined. I think that's what you're looking for here. $\endgroup$ – HDE 226868 Sep 28 '14 at 16:32
  • $\begingroup$ @HDE226868 Not exactly sure if that helps. $\endgroup$ – CoilKid Sep 28 '14 at 16:33
  • $\begingroup$ Think of the tip-to-tail method of adding vectors. The greater the magnitude, the greater the length. $\endgroup$ – HDE 226868 Sep 28 '14 at 16:34
  • $\begingroup$ Hrm. This is not homework. I'm just curious what's going on. $\endgroup$ – CoilKid Sep 28 '14 at 16:52
  • $\begingroup$ Hi CoilKid. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Sep 28 '14 at 18:53

The $x$ and $y$ velocities should not add to $V_0$. To understand why, imagine something moving with $V_x = 1 \frac{m}{s}$ and $V_y = -1 \frac{m}{s}$. This is something going horizontally and down; there's no reason why its velocity should be zero.

The answer is that $V_0$ is the length of the velocity vector $\vec{V}$, and so it's calculated using Pythagoras' theorem, like this: $V_0 = \sqrt{V_x^2 + V_y^2}$. Let's check: if $V_x = V_0 \cos \theta$ and $V_y = V_0 \sin \theta$, then $V_0 = \sqrt{V_0^2 \cos^2 \theta + V_0^2 \sin^2 \theta} = V_0\sqrt{\cos^2 \theta + \sin^2 \theta} = V_0$.

  • $\begingroup$ But if you were to shoot a projectile at exactly $45^{\circ}$ and at 10m/s the velocity should be evenly split. +5xm/s, +5ym/s. Right? $\endgroup$ – CoilKid Sep 28 '14 at 16:40
  • $\begingroup$ No. $10 \cos 45^{\circ}$ and $10 \sin 45^{\circ}$. $\endgroup$ – ProfRob Sep 28 '14 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.