How do I nicely represent the final answer?

Problem: On a flat field a canon ball with mass m is fired under an angle θ with an initial velocity of v0. Example 1.8 showed that the horizontal distance is R = v0^2 / g * sin 2θ. So the optimal angle is equal to π/4: then R = v0^2 /g. Solve the same problem in case m = 1 kg, v0 = 10 m/s, g = 9.81 m/s^2, but now including a friction force equal to: Ð→ F w = −γ∣ Ð→v ∣ Ð→v = −γvÐ→v , γ = 0.1. Find λ in θ = λπ/4 for which the distance is maximal. For every correct digit in the value of λ you score 0.1 point with a maximum of 0.5 in total. The value of λ starts with 0, for you to find p1, p2, p3, p4, p5 in λ = 0.p1p2p3p4p5, with every 0 ≤ pk ≤ 9.

My code:

v0 = 10
gamma = 0.1
for k=1:101
    theta = 0.8444 * pi/4 + (k-51)*0.0001;
    dt = 0.0000001;
    x = 0;
    y = 0;
    vx = v0 * cos(theta);
    vy = v0 * sin(theta);
    x = x + dt * vx;
    y = y + dt * vy;
    while (y>0)
        v = sqrt(vx*vx+vy*vy);
        vx = vx - dt * gamma * vx;
        vy = vy - dt * 9.81 - dt * gamma * vx;
        x = x + dt * vx;
        y = y + dt * vy;
    end;
    t(k) = theta
    a(k) = x
end;
plot(a)
[vv, jv] = max(a)
t(jv) / (pi/4)

[Bob Thompson]

What do you mean by 'nicely represent?' This seems like an asthetic question, which is very arbitrary, in my opinion.

[The Merchant]

So that i represent the final value of λ in a good way

[Bob Thompson]

That didn't really answer my question. What is wrong with having lambda just output as a variable? Do you want it output in a file? Do you want it displayed in a text message? Do you want to plot the values? Do you want to make a GUI for it? Do you want to post it to a website?

[The Merchant]

I want to post it as a string, like: The value of λ is: 0.p1p2p3p4p5etc

[Bob Thompson]

I would recommend fprintf or sprintf for that.

[The Merchant]

how would i extract teh value tho

[Bob Thompson]

Make a variable for it. I can't really do that for you because I don't really know what lambda represents for your code. I assume it's an angle, because you have theta and are using cos and sin, but that's about all I know.

[Adam Danz]

Original question by OP in case it is deleted (this user has deleted many questions after being answered).

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Problem: On a flat field a canon ball with mass m is fired under an angle θ with an initial velocity of v0. Example 1.8 showed that the horizontal distance is R = v0^2 / g * sin 2θ. So the optimal angle is equal to π/4: then R = v0^2 /g. Solve the same problem in case m = 1 kg, v0 = 10 m/s, g = 9.81 m/s^2, but now including a friction force equal to: Ð→ F w = −γ∣ Ð→v ∣ Ð→v = −γvÐ→v , γ = 0.1. Find λ in θ = λπ/4 for which the distance is maximal. For every correct digit in the value of λ you score 0.1 point with a maximum of 0.5 in total. The value of λ starts with 0, for you to find p1, p2, p3, p4, p5 in λ = 0.p1p2p3p4p5, with every 0 ≤ pk ≤ 9.

My code:

v0 = 10
gamma = 0.1
for k=1:101
    theta = 0.8444 * pi/4 + (k-51)*0.0001;
    dt = 0.0000001;
    x = 0;
    y = 0;
    vx = v0 * cos(theta);
    vy = v0 * sin(theta);
    x = x + dt * vx;
    y = y + dt * vy;
    while (y>0)
        v = sqrt(vx*vx+vy*vy);
        vx = vx - dt * gamma * vx;
        vy = vy - dt * 9.81 - dt * gamma * vx;
        x = x + dt * vx;
        y = y + dt * vy;
    end;
    t(k) = theta
    a(k) = x
end;
plot(a)
[vv, jv] = max(a)
t(jv) / (pi/4)