Physics golf: inclined shooting

Question

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.

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

Answer 1

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:

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

Answer 2

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.

Answer 3

Python3 - 65 chars

from math import*
f=lambda α,β,u:(tan(α)+tan(β))*u*u*.2*cos(β)**2