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An angry bird is shot at an angle \$β\$ to the horizontal at a speed \$u\$. The ground is steep, inclined at an angle \$α\$. Find the horizontal distance \$q\$ that the bird traveled before it hit the ground.

diagram showing an inclined plane at angle α and an angry bird being fired at angle β at speed u

Make a function \$f(α, β, u)\$ that returns the length \$q\$: the horizontal distance that the bird traveled before it hit the ground.

Constraints and notes:

  • \$-90° < α < 90°\$.
  • \$0° < β < 180°\$.
  • \$α < β\$.
  • \$0 \le u < 10^9\$.
  • Assume acceleration due to gravity \$g = 10\$.
  • You may use radians instead of degrees for \$α\$, \$β\$.
  • Dimensions of \$u\$ are irrelevant as long as they are consistent with \$g\$ and \$q\$.
  • No air resistance or anything too fancy.

Shortest code wins.

See the Wikipedia article on projectile motion for some equations.

Samples:

f(0, 45, 10) = 10
f(0, 90, 100) = 0
f(26.565, 45, 10) = 5
f(26.565, 135, 10) = 15
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  • \$\begingroup\$ As I saw some confusion about the formula, here it is for others to use it: q = ABS[1/5 u^2 Cos[β] Sec[α] Sin[β - α]] \$\endgroup\$ Feb 14, 2011 at 12:09

3 Answers 3

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Java

Works for radians only

double q(double a, double b, double u){
          return (Math.abs(((-Math.tan(a)+(Math.tan(b)))*(u*u)*(0.2*(Math.cos(b)*Math.cos(b))))));
      }

Golfed Version (Thanks to Peter)

double z=u*Math.cos(b);return(Math.tan(b)-Math.tan(a))*z*z/5;

Maths Used:

$$ q = ut\cos \beta \\ q\tan \alpha = ut\sin \beta - 0.5 \times 10 t^2 \\ - \tan \alpha + \tan \beta = 5\frac q {u^2} \sec^2 \beta \\ q = \frac {(\tan \beta - \tan \alpha)u^2} {5\sec^2 \beta } $$

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  • \$\begingroup\$ There is something wrong with this... I just cant figure out correctly, can some1 help? \$\endgroup\$ Feb 12, 2011 at 23:44
  • \$\begingroup\$ This formula is not correct. Please see comment at gnibbler's post \$\endgroup\$
    – Eelvex
    Feb 13, 2011 at 21:37
  • \$\begingroup\$ So yet, we dont have any perfect solution :) \$\endgroup\$ Feb 13, 2011 at 23:06
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    \$\begingroup\$ updated the formula... fire some testcases now please \$\endgroup\$ Feb 13, 2011 at 23:42
  • \$\begingroup\$ You can save a few chars - Math.abs is unnecessary, -x+y is shorter as y-x, *0.2 is shorter as /5, and you have unnecessary brackets. OTOH you're missing the return type of the method. \$\endgroup\$ Feb 14, 2011 at 0:06
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Haskell (37 35)

Based on Aman's solution:

q a b u=(tan a+tan b)*u*u*cos b^2/5

I think, this problem isn't real code-golf, as it is more implementing a formula than really doing some algorithm.

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  • \$\begingroup\$ Maybe you are right, since the formula is already too short. \$\endgroup\$
    – Eelvex
    Feb 12, 2011 at 23:45
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    \$\begingroup\$ Would something like /5 or /5. work? \$\endgroup\$
    – Nabb
    Feb 13, 2011 at 15:51
  • 1
    \$\begingroup\$ This formula is not correct. Please see comment at gnibbler's post. \$\endgroup\$
    – Eelvex
    Feb 13, 2011 at 21:37
2
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Python3 - 65 chars

from math import*
f=lambda α,β,u:(tan(α)+tan(β))*u*u*.2*cos(β)**2
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  • \$\begingroup\$ That's not quite correct. 1) f should always be positive and 2) for α > 0 it returns a larger value than for a=0, which is not possible. \$\endgroup\$
    – Eelvex
    Feb 13, 2011 at 21:36
  • \$\begingroup\$ Ah well, I copied FUZxxl's formula :/ \$\endgroup\$
    – gnibbler
    Feb 14, 2011 at 1:04

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