Projectile motion question totally stuck

[Delphi51]

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.

 

[emyt]

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

 

[emyt]

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

 

[Delphi51]

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.

 

[emyt]

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

 

[rl.bhat]

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.

 

[emyt]

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?

 

[rl.bhat]

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.

 

[emyt]

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

 

[rl.bhat]

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.

 

[emyt]

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

 

[emyt]

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

 

[emyt]

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

 

[Delphi51]

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.

 

[emyt]

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?

 

[Delphi51]

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 . . .

 

[emyt]

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

 

[Delphi51]

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.

 

[emyt]

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 :(

 

[Delphi51]

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'.

 

[rl.bhat]

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.

 

[Delphi51]

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!

 

[rl.bhat]

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.

 

[Delphi51]

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!

 

[rl.bhat]

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

 

[Delphi51]

Thanks! I haven't seen that trig identity for years. It sure is handy in this problem.

emyt, if you are still watching, he has used
sin(x+y) + sin(x-y) = 2sin(x)cos(y)
You probably did it in high school using
sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
and the same with y replaced by -y. These last identities are the commonly used "addition and subtraction formulas".