# Homework Help: Projectile motion question totally stuck

1. Oct 20, 2009

### emyt

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

v = v0 + at?

x = x0 + v0t + at^2 ?

3. 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

2. Oct 20, 2009

### 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.

3. Oct 20, 2009

### emyt

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

Vx0 = V0costheta

Vy0 = V0sintheta

Rsinalpha = V0sintheta - 1/2gt^2

Rcosalpha = V0costheta

thanks a lot

Last edited: Oct 20, 2009
4. Oct 20, 2009

### RoyalCat

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.

5. Oct 20, 2009

### emyt

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

6. Oct 20, 2009

### ConfusedPupil

Isn't the max for any equation 45 degrees

7. Oct 20, 2009

### 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.

8. Oct 20, 2009

### emyt

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

9. Oct 20, 2009

### 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.

10. Oct 20, 2009

### look416

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

11. Oct 20, 2009

### emyt

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

thanks again

12. Oct 20, 2009

### 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!

13. Oct 20, 2009

### 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

14. Oct 20, 2009

### look416

oh i got it
thx for clearing my confusion

15. Oct 20, 2009

### 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?

16. Oct 20, 2009

### emyt

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

17. Oct 20, 2009

### 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!

18. Oct 20, 2009

### emyt

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

19. Oct 20, 2009

### emyt

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

Last edited: Oct 20, 2009
20. Oct 20, 2009

### 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?

21. Oct 20, 2009

### emyt

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?

22. Oct 20, 2009

### 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.

23. Oct 20, 2009

### Jebus_Chris

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

$$y = vosin(\theta - \alpha) t -\frac{1}{2}gcos\alpha t^2$$

If 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

Last edited: Oct 20, 2009
24. Oct 20, 2009

### 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)

25. Oct 20, 2009

### Jebus_Chris

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