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 $.
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 $.
