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I was watching the MIT lecture about projectile motion and the lecturer asked why $$D=\frac{(v_0)^2\sin(2\alpha)}{g}$$ Why is it $(V_0)^2$ not $V_0$? It is a hypothetical question i know that the right answer is $(V_0)^2$.

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  • $\begingroup$ hi can i suggest you edit your question to show how you worked on it to arrive at Vo. I'm not saying which is the correct answer (because I don't know), but showing your own calculations line by line might help someone spot any mistake you might have made. regards $\endgroup$ – user74893 Mar 23 '15 at 22:53
  • $\begingroup$ I am sorry but it is my first time on the site but there is no calculations for it Vo it is hypothetical question $\endgroup$ – Radwa Adel Mar 23 '15 at 22:55
  • $\begingroup$ We use MathJaX here to render formulae. A tutorial is here. $\endgroup$ – ACuriousMind Mar 24 '15 at 0:31
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On an intuitive level, the initial speed $v_0$ can be considered to have two effects: one on the horizontal velocity, and one on the vertical; the former affects the range in a direct sense, and the latter increases the time the projectile is in the air. The combination of both of these gives an overall $v_0^2$ contribution.

If you were to increase the horizontal velocity on its own (not touching the vertical), you would proportionally increase the range. Similarly if you increased the vertical velocity you would increase the range. Increasing both of these, which is what increasing $v_0$ does, will increase the range twice, but multiplicatively. Think of it like a square - if you increase both sides by the same amount (e.g. 3 times), you increase the area by that amount squared (9 times).

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  • $\begingroup$ Yes but how the combination of both gives an overall (V_0)^2 the time has doubled and at the same time velcity has doubled so the range now is four times than before but why is it (v_0)^2 And not 2 (V_0) ... i dont mean mathematically but physicly explaining ?! $\endgroup$ – Radwa Adel Mar 23 '15 at 23:35
  • $\begingroup$ If you were to increase the horizontal velocity on its own (not touching the vertical), you would proportionally increase the range. Similarly if you increased the vertical velocity you would increase the range. Increasing both of these, which is what increasing $v_0$ does, will increase the range twice, but multiplicatively. Think of it like a square - if you increase both sides by the same amount, you increase the area by that amount squared... Try your example about doubling except replace it with tripling;you should get an increase by $9=3^2$. It doesn't help with that $2\times2 = 2^2$! $\endgroup$ – danimal Mar 23 '15 at 23:52
  • $\begingroup$ But the relation between the horizontal velocity and the vertical not like a square but traingle with 90° angle and the range isnt affected by the vertical velocity because when the vertical reaches zero at the heighest point the range doesnt stop there ? $\endgroup$ – Radwa Adel Mar 24 '15 at 12:11
  • $\begingroup$ the square is just a figurative analogy to express the fact that changing one thing (the length of the side of the square, or the initial speed) in two directions (horizontal/vertical) will combine to give a squared effect relationship $\endgroup$ – danimal Mar 24 '15 at 12:14
  • $\begingroup$ $v_0$ is the INITIAL speed. The vertical velocity is of course changing the whole way through the flight, but $v_0$ is constant, and determines the length of time the projectile is in the air $\endgroup$ – danimal Mar 24 '15 at 12:15
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When a projectile is launched at an initial velocity $V_0$ and an angle $\theta$, the projectile has both a horizontal ($x$-) component and vertical ($y$-) component, as shown below: Projectile Velocity

From the diagram, you can see that $v_x=v_0 \mathrm{cos\theta}$ and $v_y=v_0 \mathrm{sin\theta}$.

If we try to solve for the horizontal range of your projectile, we can use the equation $$v_x=\frac{\Delta x}{\Delta t}$$ where $\Delta x$ is the horizontal distance traveled by the projectile and $\Delta t$ is the time the projectile spends in air. If we rewrite $v_x$ in terms of the initial velocity $v_0$ we get that $$v_0 \mathrm{cos\theta}=\frac{\Delta x}{\Delta t}$$. If we clean up the expression a little and solve for the range, $\Delta x$ we get that $$\Delta x= \Delta t v_0 \mathrm{cos\theta}$$. The problem, though, is that we don't know how long the projectile is in air. Fortunately, we can determine the time the projectile spends in air by looking just at the vertical motion of of the projectile. Because the projectile accelerates while in air, due to gravity, the motion in the vertical direction is given by $$\Delta y= \frac{1}{2}g(\Delta t)^2 +v_{0y}\Delta t$$ where $\Delta y$ is the vertical displacement of the projectile, $v_{0y}$ is the initial velocity in the $y$-direction, and $\Delta t$ is the same time interval that we saw in the equation for motion in the $x$-direction. If the projectile is launched on a level surface (which you have assumed, but not stated, in the equation in your question) then $\Delta y=0$. Additionally, $v_{0y}$ can be given by $v_y=v_0 \mathrm{sin\theta}$ like we saw earlier. As a result $$0=\frac{1}{2}g(\Delta t)^2+\Delta t v_0 \mathrm{sin\theta}$$ If we try to solve for $\Delta t$, we see that one trivial solution is that $\Delta t=0$. The other solution is $$\Delta t=\frac{-2v_0\mathrm{sin}\theta}{g}$$ If we substitute $\Delta t$ into our horizontal motion equation, we get that $$\Delta x= \frac{-2v_0^2 \mathrm{sin}\theta \mathrm{cos}\theta}{g}$$ At first this doesn't look like what you have, but there is a trig identity that states that $$\mathrm{sin}(2\theta)=2\mathrm{sin}(\theta)\mathrm{cos}(\theta)$$ so the above expression simplifies to $$\Delta x=\frac{-v_0^2 \mathrm{sin}(2\theta)}{g}$$ where the negative sign is just a consequence of assuming $g=-9.8m/s^2$ so that the negative sign associated with gravity is already tucked inside the symbol.

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  • $\begingroup$ It is the same exept the negative sign but because the doctor gives g=9.8 always so ay=-g ... replacing it in the equation Y_t=-1/2g(t)^2+V_0*t ... and that is the mathematical proof i know it i want the physical why it is propotional to (V_0)^2 $\endgroup$ – Radwa Adel Mar 24 '15 at 3:43
  • $\begingroup$ Information only: The original formula used sin^2 alpha and not sine alpha. Somebody corrected it along the way making the identity that you used valid. $\endgroup$ – Russell McMahon Mar 24 '15 at 12:07
  • $\begingroup$ @RadwaAdel the answer is tucked into the math: The horizontal range is given by the produce of the horizontal velocity and time. However, the time spent in air can be found by using the equation for vertical motion of the projectile. As a result, the range is determined by how fast the projectile goes horizontally, but also how fast the projectile goes vertically (because a greater $v_{0y}$ will keep the projectile in air longer. As a result, you get a $v_0^2$ term. Dimensional analysis will also show you that that works. $\endgroup$ – Sean Mar 24 '15 at 13:43
  • $\begingroup$ Yes that is what wanted to understand not the proof i mean . Thank you very much $\endgroup$ – Radwa Adel Mar 24 '15 at 13:47
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Assume that alpha is angle relative to horizontal.
Swap cos and sin below if alpha is angle wrt vertical.

Time relates to time to rise to "apogee" and return.
Standard expression for time to apogee is $t = V/g$ ... 0 This is just stated here but can be derived if desired

Time of flight T relates to initial vertical velocity = $V \times \sin(alpha)$
Total time = ascent + return time = 2 x ascent time.

$T = 2 * V/g * \sin(\alpha)$ ... 1 = 2 x '0' above

Horizontal distance D = time x horizontal velocity $D = T * V * \cos(\alpha)$ ... 2

Substitute (1) into (2), replacing T
$D = [2 * V/g * \sin(\alpha)] * [V x \cos(\alpha)]$ ... 3

$D = 2 * {V^2}/g * \sin(\alpha) * \cos(\alpha)$

This has $V^2$ as in the original.
And $\mathrm{sin}(2\alpha)=2\mathrm{sin}(\alpha)\mathrm{cos}(\alpha)$

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  • $\begingroup$ And 2 sin(alpha)*cos(alpha) =sin(2alpha) $\endgroup$ – Radwa Adel Mar 24 '15 at 12:14

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