Shell Projected up a Hill?

Let the the horizontal from a point that is at the foot of a hill is 0,0 (origin)

The line on the surface of the hill slope will be y = tan φ x

The shell will be at point x,y at any time t , so that x = v cos α t and y = v sin α t – 0.5 g t^2

we get t = x / ( v cos α)

y =( v sin α) x / ( v cos α) – 0.5 g ( x / ( v cos α) )^2

y = x tan α – 4.905 x^2 / ( v cos α)^2 this parabola will intersect y = tan φ x

so that tan φ x = x tan α – 4.905 x^2 / ( v cos α)^2

tan φ = tan α – 4.905 x / ( v cos α)^2

4.905 x / ( v cos α)^2 = tan α – tan φ

x = ( tan α – tan φ ) ( v cos α)^2 / 4.905

y = tan φ ( tan α – tan φ ) ( v cos α)^2 / 4.905

distance to where the shell reaches the ground measured up the face of the hill. = ( x^2 + y^2)^0.5=s=

=[ {( tan α – tan φ ) ( v cos α)^2 / 4.905}^2 + { tan φ ( tan α – tan φ ) ( v cos α)^2 / 4.905}^2]^0.5

ans

for max s, ds /dα = 0

This one has the interesting answer:

α = π/4 + φ/2

for maximum range, as measured on the ground up the hill.

When I get time, I’ll explain how this was arrived. It’s a nice and simple result, giving π/4 for flat ground, and π/2 (obviously!) for vertical faces.

Edit: Dr D, very nice, short and sweet derivation. I don’t think I can top that. But I will try, nonetheless.

Edit 2: If we use the generalized equation of parabolic trajectory, where s = Tan(α), which is of the form:

-(1/2)(1+s²)x² + sx

and then add an infinitesimal ∆s to s, i.e., the trajectory of a shell shot slightly steeper, the two trajectories meet at a point which is at an angle 2α – π/2, which is does lead to the same result as above. In other words, consider the family of parabolic trajectories that a shell makes shot at varying angles from 45° to 90°, which envelope forms a curve, which is another parabola (1/2)(1 – x²). The intersection of the line Tan(φ) x with this curve marks the point where the parabolic trajectory passing through it has the maximum range for the given hill slope angle φ. It is the null point.

(For those that still find this reasoning vague, first consider an arbitrary slope intersecting the envelope curve. Then any parabolic trajectory NOT passing through this intersection point will intersect the slope line at a distance less than to the original intersection point.)

Edit 3: Given v, g, the equation of the envelope curve is:

(1/2)( v²/g – (g/v²)x² )

Given φ, the distance from (0,0) to intersection with the envelope curve is:

(v²/2g) (Csc(π/4 + φ/2))² THIS IS THE ANS YOU SEEK!

Given α, the distance from (0,0) to where the trajectory parabola is tangent to the envelope curve is:

(v²/2g) (Csc(α))²

which is to be entirely expected, since α = π/4 + φ/2.

Edit 4: For v = 1 m/s, g = 9.8 m/s², φ = 20° and α = 55°, we have:

(v²/2g) (Csc(π/4 + φ/2))² = (v²/2g) (Csc(α))² = 760.352 m

See 2nd link for this specific example.

Edit 5: By differentiating Steve’s range equation, the maximum as a function of α is (v²/2g) (Csc(α))², so there is agreement.

Source(s): http://i254.photobucket.com/albums/hh120/Scythian1…

http://i254.photobucket.com/albums/hh120/Scythian1…71

Everyone used up all the good stuff.

You forced me to create a new

planet.

I’m putting my projectiles on Planet Slope

where the gravitational field is slanted

and Earthlings (including Newton) cannot

catch apples falling out of trees. This is

a simple place. There are no hills, and for

some reason they always fire shots against

the gravitational “wind”.

g = 1, v = 1

Φ = θ-φ

R = vx(θ,φ) * t(Φ,φ)

vx = v * Cosθ / Cosφ ==>v* “velocity ratio”

Flight time given the strange gravitational

field on this planet:

T = 2 * v * sinΦ / (g * cosφ)

Range on flat ground.

R = vx* T

R = [v * Cosθ / Cosφ] * [2 * v * sinΦ / (g * cosφ]

R = [v * Cosθ / Cosφ] * [2 * v * sin[ θ-φ] / (g * cosφ]

R = 2 * cosθ * sin(θ-φ) / cosφ^ 2

R = [2/cosφ^ 2] * cosθ * sin(θ-φ)

dR/dθ = [2/cosφ^ 2] * cosθ * sin(θ-φ) = 0

==>

dR/dθ = [2/cosφ^ 2] * cos(2θ-φ) =0

cos(2θ-φ) = 0

arcos[cos(2θ-φ)] = arccos[0]

2θ-φ = 90

2θ = 90 + φ

θ = α = π/4 + φ/2

Many nice answers above me.

______________________

Edit: Let’s put some numbers to it:

Launch angle = 45+20/2 = 55

a = 55 >>launch angle relative to ground

b = 20 >>hill angle

g = 9.8 m/s^2

v = 1 m/s

R = (2 * v ^ 2 * cos(a) * Sin(a – b)) / (g * cos(b) ^ 2)

R = 487 m

Change the launch angle to (+/- 5) degrees for example

and R will drop to about: 475 m

Edit:

Trying reconcile the other equations with my numbers

Using Steve’s equation I get:

R = (v ^ 2 / g)*Sin(2 * 55)*(1 – Tan(20) / Tan(55))*1/Cos(20)

R = Steve = 487m

R = 2v ^ 2 / g (Tan(55) – Tan(20)) / (Cos(20) / Cos(55) ^ 2)

R = Dr.D = 487m

Can’t seem to reconcile all?

______________________

Edit: Scythian, your last equation gives me:

D = v ^ 2 / (2 g) (1 / Sin(55)) ^ 2

= 487 m

Btw, thanks for teaching me something about

envelope curves.

Since neither Scythian or Doc D explicitly answered the first part of the question, I will chime in with:

Slant Range Rs = (V²/g)*sin(2α)*(1 – cotα*tanφ)*secφ

This one is good for any angle α & φ within the range specified in the question. Check the link below for some caveats to this equation……..

Source(s): http://en.wikipedia.org/wiki/Trajectory

REWRITTEN:

Let us work in polar cordinates. r and θ are the variables and α is the initial angle of trajectory.

y = -gsec^2 α / (2v^2) * x^2 + x* tanα

Sub y = rsinθ and x = rcosθ

tanα – tanθ = (g/2v^2) * rcosθ sec^2 α .. (1)

For a given θ = φ, we wish to find the α which maximizes r (with θ constant).

Differentiate with respect to α and set dr/dα = 0

We end up with 1 = grcosθ tanα / v^2

Sub in eqn (1) to get

tan(2α) = -1/tan(φ)

tan(2α) = -tan(90 – φ) = tan(180 – [90 – φ]) = tan(90 + φ)

Therefore 2α = 90 + φ

α = 45 + φ/2

which is the same result Scythian got.

The value of r this gives us is

r = v^2 /(g cosφ tanα)

= (v^2 /g) * sec(φ) * cot(φ/2 + 45)

which simplifies rather nicely to

r = (v^2 /g) / [sin(φ/2) + cos(φ/2)]^2

= v^2 /[g (1 + sinφ)]

**EDIT**

According to my eqn (1) we can get the distance r

r = (2v^2 /g) * (tanα – tanφ) / (cosφ sec^2 α)

After some trig manipulations, I’m finding that this answer matches with Steve’s, FGR’s, Waheed’s as well as Al P’s (except that Al P’s θ is my α).

I changed the coordinates to r and s where r is the distance along the slope and s is the perpendicular distance from the slope.

r = v cos(α – φ) t – 1/2 g sinφ t^2

s = v sin(α – φ) t – 1/2 g cosφ t^2

set s= 0

t = 2v/g sin(α – φ) / cosφ … (t = 0 is discarded)

r = v cos(α – φ) (2v/g sin(α – φ) / cosφ) – 1/2 g sinφ (2v/g sin(α – φ) / cosφ)^2

r = 2v^2/(g*cosφ) * (cos(α – φ) sin(α – φ)) – (sin(α – φ))^2 tanφ)

dr/dα = 2v^2/(gcosφ) * (cos2(α – φ) – sin2(α – φ) * tanφ)

set dr/dα = 0

0 = dr/dα = (cos2(α – φ) – sin2(α – φ) * tanφ)

0 = 1 – tan2(α – φ) tanφ

α = 1/2 atan(1/tanφ) + φ

α = 1/2 (π/2 – φ) + φ

α = π/4 +1/2 φ

The max angle is independent of velocity (as it should be). It just stretches out or decreases the shape of the parabola.

*******Addendum********

I find it interesting that you have 5 very good answers and each with a unique method of solving this problem.