Projectile motion question totally stuck

[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".