Angry birds

Problem:

Solution:

Define the reference frame with the origin at the bird's initial position, right is positive, up is positive.

Let the initial velocity of the bird to be $\vec{v}$ and the launching angle be $\theta$.


$\begin{eqnarray*} v_x(t) &=& |\vec{v}|\cos\theta \\ v_y(t) &=& |\vec{v}|\sin\theta - gt \\ x(t) &=& (|\vec{v}|\cos\theta) t \\ y(t) &=& (|\vec{v}|\sin\theta) t - \frac{gt^2}{2} + 10 \\ \end{eqnarray*}$

Now put in the end conditions and simplify

$\begin{eqnarray*} 12 = y(2.5) &=& (|\vec{v}|\sin\theta) (2.5) - \frac{g(2.5)^2}{2} + 10 \\ 13.3 &=& (|\vec{v}|\sin\theta) \\ 55 = x(2.5) &=& (|\vec{v}|\cos\theta) (2.5) \\ 22 &=& (|\vec{v}|\cos\theta) \\ \end{eqnarray*}$

Now we get $\tan(\theta) = 13.3/22$, or $\theta = 31.15^o$.