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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?

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    $\begingroup$ 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
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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 $V_0 = \sqrt{V_0^2 \cos^2 \theta + V_0^2 \sin^2 \theta} = V_0\sqrt{\cos^2 \theta + \sin^2 \theta} = V_0$.

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  • $\begingroup$ 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 $10 \sin 45^{\circ}$. $\endgroup$ – Rob Jeffries Sep 28 '14 at 16:56

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