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A particle is projected up a plane inclined at an angle $b$ to the horizontal, the angle of projection being $a$ above the horizontal.

Part 1: If the initial speed of projection is $V$ m/s show that range, $C$, on the inclined plane is given by: $$C= 2 V^2 \frac{\sin(a-b) \cos(a)}{g\cos(b)^2}$$

Part 2: Determine the maximum range ($C$ max) on the plane and the value of $b$ for which this occurs. maximising the trigonometric function is the main problem here

Here is a diagram:

enter image description here

Jim
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Jim
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    $C$ is unbounded because cos can be made infinitessimally small. – Suzu Hirose Nov 12 '14 at 08:24
  • Even if the angles have to be in the range 0 to 90 degrees, the cosine can be made infinitesimally small. $cos(90-\epsilon)\approx\epsilon$. – Suzu Hirose Nov 12 '14 at 08:32
  • This actually came from a question about a particle being projected up an inclined plane (if that helps). The question says: A particle is projected up a plane inclined at an angle b to the horizontal, the angle of projection being a above the horizontal. part 1: if the initial speed of projection is V m/s show that range, C, on the inclined plane is given by: C= 2V^2sin(a-b)cos(a)/(cos(b)^2) part 2: determine the maximum range (R max) on the plane and the value of b for which this occurs. – Jim Nov 12 '14 at 08:34
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    sorry im not sure how to post a diagram. its just a 2 dimensional model of a projectile being fired onto a ramp – Jim Nov 12 '14 at 09:14
  • sorry i wasnt really precise saying 'algebraically' analysis is what i mean – Jim Nov 12 '14 at 09:15
  • I am such an idiot - very important is to say that a must be greater than b, because the particle must land on the plane, the angle of projection has to greater than the angle of the ramp sorry for all confusion created. thanks for all responses so far. – Jim Nov 12 '14 at 09:17
  • Im sorry, C and R are the same thing. it was a typing error, please excuse me. Range is defined as the distance along the inclined plane between the base of the plane and the point where the projectile lands on the plane. – Jim Nov 12 '14 at 12:15
  • thank you for making the original post easier to understand – Jim Nov 12 '14 at 12:23
  • This is a fairly straightforward problem where you can use the horizontal & vertical displacements at time t – Yaitzme Nov 12 '14 at 12:32
  • @Jim thanks for the diagram. You can unclutter this by deleting outdated comments using the tiny "x". – mvw Nov 12 '14 at 12:33
  • @Yaitzme Yes, but I at least needed a diagram to understand the problem. Now it is just a matter of physics. See how late that not unimportant gravitational acceleration $g$ entered the question? – mvw Nov 12 '14 at 12:36
  • just to clarify, part 1 I understand, what I'm interested in is how to find the relationship between a and b that gives greatest C, (first equation) and then apply this to get the max 'range' for a given v. thanks so much for responses – Jim Nov 12 '14 at 12:38
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    I think $a$ and $b$ are reversed on the figure. You want $b,$ not $a,$ to be the angle between the inclined plane and horizontal, is that right? – David K Nov 12 '14 at 17:32

2 Answers2

1

Here is a diagram of the problem with a few helpful elements added:

The figure illustrates the fact that if not for the influence of gravity, the projectile would simply travel a distance $Vt$ along the line at an angle $a$ above the horizontal in a certain time $t,$ but because of gravity, the projectile is deflected downward a distance $\frac 12 gt^2,$ so it ends up at the intersection of the blue parabola and the vertical red line. This point is the landing point on the inclined surface if we choose $t$ to be the amount of time the projectile is in flight before striking that surface.

Now we do a little trigonometry. We draw a right triangle with the segment $Vt$ as the hypotenuse and an extension of segment $C$ as one of the legs. The vertical red line cuts a smaller right triangle from this triangle; the smaller triangle has hypotenuse $\frac 12 gt^2$ and angle $b$ at the topmost vertex, hence the adjacent leg is $\frac 12 gt^2 \cos b.$ But that leg is also the leg opposite the angle $a-b$ of the larger right triangle, so we have $$Vt \sin (a-b) = \frac 12 gt^2 \cos b.$$

Now consider the horizontal segment in the figure, which is the leg adjacent to angle $b$ in a right triangle whose hypotenuse is $C,$ so its length is $C \cos b,$ but it is also the leg adjacent to angle $a$ in a right triangle whose hypontenuse is $Vt.$ This gives us $$C \cos b = Vt \cos a.$$

From these two equations we can conclude that $$\left(\frac 12 gt^2 \cos b\right)(C \cos b) = (Vt \sin (a-b))(Vt \cos a).$$

We have already assumed that $t > 0,$ so we can cancel factors of $t.$ Assuming $b$ is less than a right angle, $\cos b > 0$ as well, and we can assume $g > 0.$ We can use these facts to isolate $C$ on one side of the equation. The result is $$C = \frac{2V^2 \sin (a-b) \cos a}{g \cos^2 b}.$$

The second part of the problem is to maximize $C.$ At least one version of the posted question allows us to vary $b$ in order to maximize $C.$ But we can make $C$ as large as we want merely by keeping $a$ constant and letting $b$ approach $-\frac\pi2$ radians--that is, incline the plane as steeply downwards as we like, but never quite vertical (because then there is no second intersection of the trajectory with the plane, and $C$ is undefined). We can even allow $a$ to vary and still $C$ is unbounded as long as $a$ is not too close to $b.$

So I will answer a more interesting question, which I suspect is what was meant to be asked: find $a$ that maximizes $C$ for given constant values of $V,\ g,$ and $b.$

To solve this, observe the trigonometric identity $$\sin x - \sin y = 2 \sin\left(\frac{x-y}{2}\right) \cos\left(\frac{x+y}{2}\right).$$ Conveniently, there is a product of a sine and cosine in the quantity we want to maximize. So we would like to have $$\begin{eqnarray} a - b &=& \frac12(x-y), \\ a &=& \frac12(x+y). \end{eqnarray}$$ Solving for $x$ and $y$ in terms of $a$ and $b,$ we find that $$\begin{eqnarray} x &=& 2a - b, \\ y &=& b. \end{eqnarray}$$ Therefore, we can use the identity to substitute for $\sin (a-b) \cos a$ as follows: $$C = \frac{2V^2}{g \cos^2 b} (\sin (2a-b) - \sin b).$$

But according to the conditions of the problem, everything on the right-hand side of this equation is a constant except for $a.$ We therefore maximize $C$ by maximizing $\sin (2a-b),$ which occurs when $2a-b$ is a right angle. Measuring in radians, that gives us $$a = \frac12\left(\frac\pi2 + b\right).$$ Measured in degrees, $a = 45 + \frac12 b.$

David K
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Here's me having a go at it:

Now, let the particle hit the plane after time t.

Horizontal Displacement = $C*\cos b$
Vertical Displacement = $C*\sin b$

Horizontal velocity = $V*\cos a$ (along positive x-axis)
Vertical velocity = $V*\sin a$ (along positive y-axis)

After time t:

$V*\cos a*t = C*\cos b$ . . . (1)
$V*\sin a *t - 0.5*g*t^2 = C*\sin b$ . . . (2)

From (1) obtain the value of $t$ and plug it into (2) and rearrange to answer Qn 1

Senex Ægypti Parvi
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Yaitzme
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