Projectile motion question totally stuck

[emyt]

Homework Statement

A cannon is arrange to fire projectiles, with initial speed V0, directly up the face of a hill of elevation angle alpha. At what angle from the horizontal should the cannon be aimed to obtain the maximum possible range R up the face of the hill?

Homework Equations

v = v0 + at?

x = x0 + v0t + at^2 ?

The Attempt at a Solution

I have no idea what to do, please help my self esteem is totally shot .. I can't believe that I actually have no clue

 

[Delphi51]

Just begin as you would any projectile problem.
Separate the initial velocity into horizontal and vertical parts. Then write two headings:
Horizontal and Vertical. Decide in each case whether you have constant speed or accelerated motion and write the appropriate formula(s).
You'll need some model for the hill - it is like a straight line on a y vs x graph so you know its equation.

 

[emyt]

Just begin as you would any projectile problem.
Separate the initial velocity into horizontal and vertical parts. Then write two headings:
Horizontal and Vertical. Decide in each case whether you have constant speed or accelerated motion and write the appropriate formula(s).
You'll need some model for the hill - it is like a straight line on a y vs x graph so you know its equation.

Hi, thanks for the reply - I tried doing something like..

Vx0 = V0costheta

Vy0 = V0sintheta

Rsinalpha = V0sintheta - 1/2gt^2

Rcosalpha = V0costheta
thanks a lot

 

[RoyalCat]

Hi, thanks for the reply - I tried doing something like..

Vx0 = V0costheta

Vy0 = V0sintheta

Rsinalpha = V0sintheta - 1/2gt^2

Rcosalpha = V0costheta

but this didn't get me anywhere.. are you telling me to treat the hill as if it were a flat surface? what do I do with the angle the hill makes then?

thanks a lot

Realign your coordinate system so that the positive x-axis is pointed up the slope. The angle of the slope comes into play when you calculate the accelerations in the x and y axes.

 

[emyt]

Realign your coordinate system so that the positive x-axis is pointed up the slope. The angle of the slope comes into play when you calculate the accelerations in the x and y axes.

can I just consider the hill as a straight line, calculate the angle from there and then just add on alpha after? I thought of this but I wasn't sure..

thanks

 

[ConfusedPupil]

Isn't the max for any equation 45 degrees

 

[Delphi51]

45 degrees is the optimum angle to maximize the range on a flat surface. Throwing up a hill at angle alpha, the optimum angle will vary with alpha.

emyt, I was thinking that a hill of angle alpha will have a slope of tan(alpha) so its equation will be y = mx = tan(alpha)*x. [1]

For the horizontal part, constant speed, you'll have something like
x = v*cos(θ)*t [2]
For the vertical part, instead of "Rsinalpha = V0sintheta - 1/2gt^2 " I would just write y = v*sin(θ)*t - 1/2gt^2 in this notation. [3]
Sub [1] into [2] or [3] so you are looking at the situation when the projectile hits the hill. Study the remaining two equations and see if you can see what angle gives the maximum x or y that you are looking for.

 

[emyt]

45 degrees is the optimum angle to maximize the range on a flat surface. Throwing up a hill at angle alpha, the optimum angle will vary with alpha.

emyt, I was thinking that a hill of angle alpha will have a slope of tan(alpha) so its equation will be y = mx = tan(alpha)*x. [1]

For the horizontal part, constant speed, you'll have something like
x = v*cos(θ)*t [2]
For the vertical part, instead of "Rsinalpha = V0sintheta - 1/2gt^2 " I would just write y = v*sin(θ)*t - 1/2gt^2 in this notation. [3]
Sub [1] into [2] or [3] so you are looking at the situation when the projectile hits the hill. Study the remaining two equations and see if you can see what angle gives the maximum x or y that you are looking for.

hi, thanks for the reply:

I thought that I was looking for rcosalpha and rsinalpha as my final x and y.. since those would be the coordinates at the end of the range?

thanks

 

[Delphi51]

Oh, I see it now! I was confused at the lack of x's and y's. You are saying the same thing I did and you have already got it down to two equations. Missing a couple of t's, though. It's x = v*t and
y = v*t + .5*a*t^2.

If you solve the easier one for t and sub in the other, you get a nice quadratic in r so you should be able to get its maximum either using a derivative or recalling a quadratic's max from high school math.

 

[look416]

well, to find a max range
cant we use R=u²sin2\theta/g

 

[emyt]

Oh, I see it now! I was confused at the lack of x's and y's. You are saying the same thing I did and you have already got it down to two equations. Missing a couple of t's, though. It's x = v*t and
y = v*t + .5*a*t^2.

If you solve the easier one for t and sub in the other, you get a nice quadratic in r so you should be able to get its maximum either using a derivative or recalling a quadratic's max from high school math.

yeah I got it, thanks - I was repulsed by the hideous substituting... it wasn't so bad

thanks again

 

[Delphi51]

That's a powerful little formula, look416!
But it is only for shooting on a horizontal surface and doesn't apply to this problem.

Congrats, emyt!

 

[emyt]

hi, I tried putting -4.9(rcosalpha/V0x costheta)^2 + v0sintheta(rcosalpha/V0x costheta) - Rsinalpha = 0 into the quadratic formula and it got really messy :S should I be doing it like this? there are too many unknowns..

thank you

 

[look416]

oh i got it
thx for clearing my confusion

 

[Delphi51]

I didn't get such a messy quadratic as you. Mine had no c term so I could common factor it. However, I ended up with a formula for tan θ that blows up when alpha equals zero, so I must have an error.

Perhaps if we work together on it we can find each other's errors.
You are supposed to take the lead. How about we use A for alpha and T for theta so it isn't quite so messy? Do you have
r*cos(A) = v*cos(T)*t and r*sin(A) = V*sin(T)*t - .5*g*t^2
as the starting point? What did you get next?

 

[emyt]

I didn't get such a messy quadratic as you. Mine had no c term so I could common factor it. However, I ended up with a formula for tan θ that blows up when alpha equals zero, so I must have an error.

Perhaps if we work together on it we can find each other's errors.
You are supposed to take the lead. How about we use A for alpha and T for theta so it isn't quite so messy? Do you have
r*cos(A) = v*cos(T)*t and r*sin(A) = V*sin(T)*t - .5*g*t^2
as the starting point? What did you get next?

hi, I had the same thing, when I solved for t I got rcos(A)/vcos(T)

then I plugged it into the sin equation and subtracted V*sin(T) from both sides to equate it to zero so I could find rcos(A)/vcos(T) and I thought that if I could assign a number to that, I could solve for theta.. but it didn't work

thanks for working with me

 

[Delphi51]

Okay so putting t = rcos(A)/vcos(T) into r*sin(A) = V*sin(T)*t - .5*g*t^2
gives you
r*sin(A) = V*sin(T)*rcos(A)/vcos(T) - .5*g*r^2/v^2*cos^2(A)/cos^2(T)
Why now subtract V*sin(T) from both sides? Suggest we divide both sides by r and take the two terms that have no r to the left side.

This is fun - let's keep going!

 

[emyt]

Okay so putting t = rcos(A)/vcos(T) into r*sin(A) = V*sin(T)*t - .5*g*t^2
gives you
r*sin(A) = V*sin(T)*rcos(A)/vcos(T) - .5*g*r^2/v^2*cos^2(A)/cos^2(T)
Why now subtract V*sin(T) from both sides? Suggest we divide both sides by r and take the two terms that have no r to the left side.

This is fun - let's keep going!

I equated it to zero so I could try solving for rcos(A)/vcos(T), but I guess I could divide by r as well, I just thought it would work with that method from what you said before
thanks I'm on it

 

[emyt]

Okay so putting t = rcos(A)/vcos(T) into r*sin(A) = V*sin(T)*t - .5*g*t^2
gives you
r*sin(A) = V*sin(T)*rcos(A)/vcos(T) - .5*g*r^2/v^2*cos^2(A)/cos^2(T)
Why now subtract V*sin(T) from both sides? Suggest we divide both sides by r and take the two terms that have no r to the left side.

This is fun - let's keep going!

hi, what did you get? :S I'm doing so much expanding and whatnot it doesn't seem right

 

[Delphi51]

sin(A) = V*sin(T)*cos(A)/vcos(T) - .5*g*r/v^2*cos^2(A)/cos^2(T)
after dividing by r.
I'll take the r term to the left and the sin(A) to the right:
.5*g*r/v^2*cos^2(A)/cos^2(T) = V*sin(T)*cos(A)/vcos(T) - sin(A)
How would you solve that for r?

 

[emyt]

sin(A) = V*sin(T)*cos(A)/vcos(T) - .5*g*r/v^2*cos^2(A)/cos^2(T)
after dividing by r.
I'll take the r term to the left and the sin(A) to the right:
.5*g*r/v^2*cos^2(A)/cos^2(T) = V*sin(T)*cos(A)/vcos(T) - sin(A)
How would you solve that for r?

oh right, thanks.. then you divide and multiply to isolate r..

r = (V*sin(T)*cos(A)/vcos(T) - sin(A)*[v^2*cos^2(A)/cos^2(T)])/.5*g

but V theta and alpha are all unknown?

 

[Delphi51]

We must consider that alpha (A) is known - the specified hill angle.
Our job is to find the theta that makes r a maximum.
To avoid the mistake I made last time, we should check at this stage to see if it makes sense when alpha is zero.

 

[Jebus_Chris]

If you rotate your coordinate plane (I believe):
x = vocos(\theta - \alpha ) t - \frac{1}{2}gsin\alpha t^2

y = vosin(\theta - \alpha) t -\frac{1}{2}gcos\alpha t^2If you don't rotate your coordinate plane:
x = vocos\theta t

y = vosin\theta t - \frac{1}{2}gt^2

tan \alpha = \frac{y}{x}

R = \sqrt{x^2+y^2}\theta = Launch angle
\alpha = Angle of incline

 

[Delphi51]

Thanks, Jebus. Regret I don't understand your very first step, the
v*cos(theta)*t. It seems to me that after you rotate, your v would no longer be v but something like v*cos(alpha).

emyt, I'm not agreeing with your last step. I get
.5*g*r/v^2*cos^2(A)/cos^2(T) = V*sin(T)*cos(A)/vcos(T) - sin(A)
.5*g*r = V^2*sin(T)*cos^3(A)/cos^3(T) - v^2*sin(A)*cos^2(A)/cos^2(T)
and then
r = 2/g*V^2*sin(T)*cos^3(A)/cos^3(T) - 2/g*v^2*sin(A)*cos^2(A)/cos^2(T)

 

[Jebus_Chris]

I probably should've stated the variables. I'll edit that in :>

 

[emyt]

If you rotate your coordinate plane (I believe):
x = vocos(\theta - \alpha ) t - \frac{1}{2}gsin\alpha t^2

y = vosin(\theta - \alpha) t -\frac{1}{2}gcos\alpha t^2If you don't rotate your coordinate plane:
x = vocos\theta t

y = vosin\theta t - \frac{1}{2}gt^2

tan \alpha = \frac{y}{x}

R = \sqrt{x^2+y^2}\theta = Launch angle
\alpha = Angle of incline

hi, if I don't rotate the coordinate plane, wouldn't x and y be Rcosalpha and Rsinalpha?
and I see how you get v0cos(theta - alpha) but how did you get gsinalphat^2 .. etc ?

thanks

 

[emyt]

Thanks, Jebus. Regret I don't understand your very first step, the
v*cos(theta)*t. It seems to me that after you rotate, your v would no longer be v but something like v*cos(alpha).

emyt, I'm not agreeing with your last step. I get
.5*g*r/v^2*cos^2(A)/cos^2(T) = V*sin(T)*cos(A)/vcos(T) - sin(A)
.5*g*r = V^2*sin(T)*cos^3(A)/cos^3(T) - v^2*sin(A)*cos^2(A)/cos^2(T)
and then
r = 2/g*V^2*sin(T)*cos^3(A)/cos^3(T) - 2/g*v^2*sin(A)*cos^2(A)/cos^2(T)

I think if you do the algebra it works out to be the same thing, I just copied and pasted thing around.. anyway, don';t you think that the answer should have a bit more of a concise method? there's point where I feel like the answer isn't going anywhere...

 

[Jebus_Chris]

hi, if I don't rotate the coordinate plane, wouldn't x and y be Rcosalpha and Rsinalpha?

thanks

Yep. That may require more manipulation of equations in the long run though.

 

[Delphi51]

In the rotated coordinates, how do you get the time and x value when it hits the hill?

 

[emyt]

Thanks, Jebus. Regret I don't understand your very first step, the
v*cos(theta)*t. It seems to me that after you rotate, your v would no longer be v but something like v*cos(alpha).

emyt, I'm not agreeing with your last step. I get
.5*g*r/v^2*cos^2(A)/cos^2(T) = V*sin(T)*cos(A)/vcos(T) - sin(A)
.5*g*r = V^2*sin(T)*cos^3(A)/cos^3(T) - v^2*sin(A)*cos^2(A)/cos^2(T)
and then
r = 2/g*V^2*sin(T)*cos^3(A)/cos^3(T) - 2/g*v^2*sin(A)*cos^2(A)/cos^2(T)

were you planning to plug that in somewhere? that's humongous

 

[Delphi51]

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.

 

[emyt]

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

 

[RoyalCat]

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.

 

[emyt]

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?

 

[emyt]

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

 

[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?