Projectile motion question totally stuck

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SUMMARY

The discussion centers on solving a projectile motion problem involving a cannon firing projectiles at an angle up a hill with elevation angle alpha. Participants emphasize the need to separate the initial velocity into horizontal (Vx0 = V0cos(theta)) and vertical (Vy0 = V0sin(theta)) components. The optimal launch angle for maximum range varies with the hill's angle, deviating from the standard 45 degrees used for flat surfaces. Key equations include the projectile's horizontal and vertical motion equations, with participants exploring methods to derive the angle that maximizes the range up the incline.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions (sine, cosine)
  • Ability to manipulate quadratic equations
  • Knowledge of coordinate systems and transformations
NEXT STEPS
  • Study the derivation of projectile motion equations on inclined planes
  • Learn about coordinate transformations in physics problems
  • Explore optimization techniques for quadratic functions
  • Research the effects of varying launch angles on projectile range
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics and projectile motion, as well as anyone looking to enhance their problem-solving skills in kinematics.

  • #31
Found a mistake.
r = 2v^2/(g*cos^2(A)) times this trig expression:
cos(A)*sin(T)*cos(T) - sin(A)*cos^2(T)
Only this last expression varies with T, so only it matters.
To get the max for r, we differentiate this with respect to T and set it equal to zero. Almost there! I already checked when A=0 and got the 45 degrees we would expect!

Jebut, your rotated solution gets as messy as ours when you set y = 0 to be on the hill, solve for t and sub into the x equation.
 
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  • #32
Delphi51 said:
Found a mistake.
r = 2v^2/(g*cos^2(A)) times this trig expression:
cos(A)*sin(T)*cos(T) - sin(A)*cos^2(T)
Only this last expression varies with T, so only it matters.
To get the max for r, we differentiate this with respect to T and set it equal to zero. Almost there! I already checked when A=0 and got the 45 degrees we would expect!

Jebut, your rotated solution gets as messy as ours when you set y = 0 to be on the hill, solve for t and sub into the x equation.

okay, thanks for checking that it works. let me check it out

so you got r to be 2v^2/(g*cos^2(A)) and which equation did you plug it into? either one of the x or y?

thanks a lot
 
  • #33
Well, if we forgo rotating, I think the best approach here would be to look at these two equations as intersecting curves/lines.

One is the regular quadratic for projectile motion, and the other would be the straight line y=\tan{\alpha}x

Setting the two to be equal, and solving for the x where they intersect will give us the x coordinate of their intersection (And the trivial solution x=0)

Since that's a linear function, maximizing x_{impact} would maximize the whole of the function. All we've got left to do from there is to differentiate with respect to \theta and equate to zero.

Now it's just a question of getting over the algebra.
 
  • #34
RoyalCat said:
Well, if we forgo rotating, I think the best approach here would be to look at these two equations as intersecting curves/lines.

One is the regular quadratic for projectile motion, and the other would be the straight line y=\tan{\alpha}x

Setting the two to be equal, and solving for the x where they intersect will give us the x coordinate of their intersection (And the trivial solution x=0)

Since that's a linear function, maximizing x_{impact} would maximize the whole of the function. All we've got left to do from there is to differentiate with respect to \theta and equate to zero.

Now it's just a question of getting over the algebra.

thanks, I was really thinking about the axis rotation.. I don't get why the answer isn't just pi/4 + alpha?

on a straight line, the maximum angle is pi/4 so if you turned the axis, it would be pi/4 + alpha?
 
  • #35
emyt said:
thanks, I was really thinking about the axis rotation.. I don't get why the answer isn't just pi/4 + alpha?

on a straight line, the maximum angle is pi/4 so if you turned the axis, it would be pi/4 + alpha?

anybody? I actually think this is a good idea, but where have I gone wrong?

thanks
 
  • #36
It just isn't that simple. In the rotated system you have accelerated motion in both directions. Consider the case when alpha is 80 degrees. Your answer says shoot at pi/4 + alpha = 45+80 = 125 degrees, which is actually away from the hill and would never hit it.

RoyalCat's idea of using x instead of r seems to reduce the algebra considerably. I worked it through to the end and got the same answer I got last night with the r. Unfortunately there seems to be an error in my work. I get tan(2θ) = - 1/tan(alpha) with a condition alpha not equal to zero. So when alpha is 10 degrees, it says shoot at -40 degrees.

Edit: I used a spreadsheet to take a good look at my solution
x = 2*v^2/g(sinθcosθ-tanα*cosθcosθ)
before taking the derivative to get the max, and it works. With alpha 10 degrees, shooting at 48 degrees gives the max range. So I just have an error in the derivative or last bit of algebra.

Recommend you use y = tan(alpha)*x to eliminate y in the original equation y = v*sinθ*t -.5*g*t^2 and use x = v*cosθ*t to eliminate t. Solve for x. About 3 lines on you'll see my x = expression above.
 
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  • #37
Delphi51 said:
It just isn't that simple. In the rotated system you have accelerated motion in both directions. Consider the case when alpha is 80 degrees. Your answer says shoot at pi/4 + alpha = 45+80 = 125 degrees, which is actually away from the hill and would never hit it.

RoyalCat's idea of using x instead of r seems to reduce the algebra considerably. I worked it through to the end and got the same answer I got last night with the r. Unfortunately there seems to be an error in my work. I get tan(2θ) = - 1/tan(alpha) with a condition alpha not equal to zero. So when alpha is 10 degrees, it says shoot at -40 degrees.

thanks for replying, I get the idea behind RoyalCat's idea, I think it's nice. I'm not incredibly familiar with the applications of calculus yet so I need some time before I can intuitively use them freely.. I'd like to try solving it with the rotational method though because it seems like an interesting route, I see what you mean about the acceleration though, I'll see how I can fix it

thanks guys
 
  • #38
how did you get x = Vcos(T-A) - 1/2g cos(A) t^2? the -1/2gcos(A) part I mean I would think that the acceleration in my new axes would be a vector that is a sum of two multiples of vectors going along the new x and y axis?

thanks
 
  • #39
Mystery solved on the tan(2θ) = - 1/tan(alpha).
When alpha = 10 degrees, you get 2θ = -80 degrees OR 100 degrees.
I should have ignored the negative solution!
So θ = 50 degrees. Shoot at 50 degrees to maximize range on the 10 degree hill.
 
  • #40
Delphi51 said:
Mystery solved on the tan(2θ) = - 1/tan(alpha).
When alpha = 10 degrees, you get 2θ = -80 degrees OR 100 degrees.
I should have ignored the negative solution!
So θ = 50 degrees. Shoot at 50 degrees to maximize range on the 10 degree hill.

yes, that is correct - the book says pi/4 + alpha/2

anyway, could you please shed some light on the rotating axis method? how come you have to multiply -1/2g by cos(A)?

x = V*cos(T - A) - 1/2gcos(A)t^2?

thanks
 
  • #41
If you call the axis along the inclined plane to be x axis, the velocity makes an angle ( θ-α).
The time of flight T = 2*v*sin( θ-α)/g*cosα.
Horizontal distance x' = v*cosθ*Τ = 2v^2*sin(θ-α)*cosθ/gcosα
R' = x'/cosα = 2v^2*sin(θ-α)*cosθ/gcos^2(α)
= [2v^2/g*cos^2(α)]*1/2[sin(2θ-α) - sinα]
The range R' on the inclined plane will be maximum when 2θ - α = π/2. Or
θ = π/4 + α/2.
 
  • #42
rl.bhat said:
If you call the axis along the inclined plane to be x axis, the velocity makes an angle ( θ-α).
The time of flight T = 2*v*sin( θ-α)/g*cosα.
Horizontal distance x' = v*cosθ*Τ = 2v^2*sin(θ-α)*cosθ/gcosα
R' = x'/cosα = 2v^2*sin(θ-α)*cosθ/gcos^2(α)
= [2v^2/g*cos^2(α)]*1/2[sin(2θ-α) - sinα]
The range R' on the inclined plane will be maximum when 2θ - α = π/2. Or
θ = π/4 + α/2.
thanks, but could you tell me why you multiplied the acceleration by cos alpha?
 
  • #43
If you rotate the axis, you have to change the acceleration. Because in projectile motion the acceleration is perpendicular to x-axis. So gcosα is perpendicular to the new x-axis.
 
  • #44
rl.bhat said:
If you rotate the axis, you have to change the acceleration. Because in projectile motion the acceleration is perpendicular to x-axis. So gcosα is perpendicular to the new x-axis.

hi, thanks.. but why how did you know to use gcosa? I would think that you should take |a| multiplied by some unit vector along the direction? but cosa is just some scalar, how does that work?

thanks
 
  • #45
Because you have rotated the original axis to new axis through an angle α, you have to take the component of g in that direction to study the motion of the projectile with respect to the new axis.
 
  • #46
rl.bhat said:
Because you have rotated the original axis to new axis through an angle α, you have to take the component of g in that direction to study the motion of the projectile with respect to the new axis.

okay.. I think I'm starting to see it.. the x component of the acceleration vector gravity in the new axis is parallel to sin alpha and the y component is parallel to cos alpha.. multiplying it by g gives you the magnitude of gravity...

man, I need to get a new textbook, NONE of this is in my textbook.

thanks
 
  • #47
emyt said:
okay.. I think I'm starting to see it.. the x component of the acceleration vector gravity in the new axis is parallel to sin alpha and the y component is parallel to cos alpha.. multiplying it by g gives you the magnitude of gravity...

man, I need to get a new textbook, NONE of this is in my textbook.

thanks

is this right?

thanks
 
  • #48
rl.bhat said:
If you call the axis along the inclined plane to be x axis, the velocity makes an angle ( θ-α).
The time of flight T = 2*v*sin( θ-α)/g*cosα.
Horizontal distance x' = v*cosθ*Τ = 2v^2*sin(θ-α)*cosθ/gcosα
R' = x'/cosα = 2v^2*sin(θ-α)*cosθ/gcos^2(α)
= [2v^2/g*cos^2(α)]*1/2[sin(2θ-α) - sinα]
The range R' on the inclined plane will be maximum when 2θ - α = π/2. Or
θ = π/4 + α/2.

never mind, I'm just trying to figure the rest out now.. I don't know why you did x'/cosa
 
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  • #49
V*costheta*T is the horizontal distance. The range he is using in the rotated coordinates is the distance along the hill.

θ = π/4 + α/2 is such a nice solution! After seeing it I figured out how to get it from my tan(2θ) = -1/tan(α) = -sin(α)/cos(α)
tan(2θ) = -sin(π/2+α)/-cos(π/2+α) = tan(π/2+α)
so 2θ = π/2+α and θ = π/4+α/2.
Pretty tricky trigonometry in the unrotated coordinates.
I didn't work the rotated on all the way through to see if it is easier.
 
  • #50
Delphi51 said:
V*costheta*T is the horizontal distance. The range he is using in the rotated coordinates is the distance along the hill.

θ = π/4 + α/2 is such a nice solution! After seeing it I figured out how to get it from my tan(2θ) = -1/tan(α) = -sin(α)/cos(α)
tan(2θ) = -sin(π/2+α)/-cos(π/2+α) = tan(π/2+α)
so 2θ = π/2+α and θ = π/4+α/2.
Pretty tricky trigonometry in the unrotated coordinates.
I didn't work the rotated on all the way through to see if it is easier.

how do you get the distance along the hill by x'/cosa?
 
  • #51
Hmmm - sure looks like an error. x' itself should be the distance along the hill. Doesn't change the final answer, though.

I should work through the rotated solution just for my late education . . .
 
  • #52
Delphi51 said:
Hmmm - sure looks like an error. x' itself should be the distance along the hill. Doesn't change the final answer, though.

I should work through the rotated solution just for my late education . . .

I'm not sure if it's an error , I don't think two totally different things can have the same answer.. hopefully, he'll come back

this question just makes me feel like quitting physics :cry:
 
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  • #53
Well, multiplying the range by cos(alpha) changes the range, but it doesn't affect the theta that gives the maximum range.

I had a go at the rotated coords. I understood Jebus' starting equations for x' and y' in post #23. But I do not understand rl.bhat's starting equation: x' = v*cosθ*Τ. Surely there is an acceleration in the x' direction as Jebus had! And Jebus' equations make for a solution at least as complicated as the ones I did in the unrotated coordinates. I'd still say the unrotated approach, maximizing x instead of r, is the easiest way to do it, though there is some tricky trig to get that nice final answer.

Don't let it depress you about physics - you'll be MUCH better at algebra, etc. in a year and this will seem easy. Don't wear yourself out on it now.
 
  • #54
Delphi51 said:
Well, multiplying the range by cos(alpha) changes the range, but it doesn't affect the theta that gives the maximum range.

I had a go at the rotated coords. I understood Jebus' starting equations for x' and y' in post #23. But I do not understand rl.bhat's starting equation: x' = v*cosθ*Τ. Surely there is an acceleration in the x' direction as Jebus had! And Jebus' equations make for a solution at least as complicated as the ones I did in the unrotated coordinates. I'd still say the unrotated approach, maximizing x instead of r, is the easiest way to do it, though there is some tricky trig to get that nice final answer.

Don't let it depress you about physics - you'll be MUCH better at algebra, etc. in a year and this will seem easy. Don't wear yourself out on it now.

it looks like he found the time it takes to go from y0 = 0 to y = 0 in the rotated coordinates, then for the original coordinates, he plugged the time back into see the total horizontal distance.

I see to be doing fairly well in mathematics, it's just something about physics that makes me feel disabled :(
 
  • #55
I did see that y' = 0 to get t = 2v/g*sin(θ-A)/cos(A).
But x' = v*cosθ*Τ should be x' = vcos(θ-A)*t - .5g*sin(A)*t^2
= 2v^2/g[cos(θ-A)sin(θ-A)/cos(A) - tan(A)sin^2(θ-A)]

Just as complicated as the unrotated expression:
x = 2v^2/g[sin(θ)cos(θ) - tan(A)cos^2(θ)]
I'd rather do dx/dθ on the x than on the x'.
 
  • #56
Sorry guys. I will clear your doubts. I have taken rotated axis as x-axis.
If O is the starting point, OX = R' is the range along the rotated axis. I have used R' because it is not the actual range.
Its projection on the horizontal axis is OX' = R'cosα. = v*cosθ*T. It is the usual way to find the distance along the horizontal axis in a projectile motion.
I am not able to draw the figures and post it. Sorry.
 
  • #57
Thanks for coming back, rl.bhat! You have a wonderfully short solution that we don't understand. The R' business is minor, but I don't understand that either. Is R' the same as x', the distance from start to finish along the hill? Or is it x, the unrotated x coordinate at the point where it strikes the hill? Or something else?

What about the x' = v*cosθ*Τ you started with?
Shouldn't it be x' = vcos(θ-A)*t - .5g*sin(A)*t^2 due to the component of mg in the x' direction?
I realize your solution must be correct because you got the correct answer, but don't understand its starting point!
 
  • #58
Draw OX,the horizontal axis, OX' along the slope of the hill making an angle α with OX. Let OP be the range R' along OX'. Draw PM perpendicular to OX. OM is the x co-ordinate of the particle when it hits the hill. x = v*cosθ*Τ= OP*cosα ( In the previous post I have mentioned it as x'). T is the time taken by the particle to reach the target. I think the rest of the things are clear.
 
  • #59
Okay, that's cleared up: x = v*cosθ*Τ, not x'.
I'm thinking you started with these equations in the rotated coordinate system:
x' = vcos(θ-α)*t - .5g*sin(α)*t^2, and y'= vsin(θ-α)*t-.5g*cos(α)*t^2
Set y' = 0 and solve for t to get the time of flight:
T = 2v/g*sin(θ-α)/cos(α)
Then so cleverly switch over to the unrotated coordinate system with
x = v*cos(θ)*t and sub that T value from 2 lines up to get
x = 2v^2/(g*cos(α))*cos(θ)*sin(θ-α)
Differentiate with respect to θ and set equal to zero to maximize:
dx/dθ = constants*[-sinθsin(θ-α) + cosθcos(θ-α)] = 0
sinθsin(θ-α) = cosθcos(θ-α)
sinθ/cosθ = cos(θ-α)/sin(θ-α)
and I'm not clear on how you got θ = π/4 + α/2 from that.
Likely I have made a mistake somewhere!
 
  • #60
x = 2v^2/(g*cos(α))*cos(θ)*sin(θ-α)
Using trig. formula of conversion of product to sum, I got
x = 2v^2/(g*cos(α))*1/2[sin(2θ-α) - sinα]
When R is max, x is maximum.
x is maximum when [sin(2θ-α) - sinα] is maximum. Sinα is constant. So sin(2θ-α) should be maximum. That means 2θ-α = π/2. Or θ = π/4 + α/2
From your calculations
dx/dθ = constants*[-sinθsin(θ-α) + cosθcos(θ-α)] = 0
[-sinθsin(θ-α) + cosθcos(θ-α)] =0
Or cos(2θ -α) = 0 ...> 2θ - α = π/2
 
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