The hunter and the monkey

If the monkey simply lets go, it will be hit by the dart. But if he jumps with sufficiently large velocity component perpendiular to the line of fire, the dart will miss.

To analyze this, consider that the motions of monkey and dart are described as $\mathbf{x}_m = \mathbf{x}_{m,0} + \mathbf{v}_{m,0} t + \tfrac12 \mathbf{g} t^2; \\ \mathbf{x}_d = \mathbf{x}_{d,0} + \mathbf{v}_{d,0} t + \tfrac12 \mathbf{g} t^2.$ Moreover, $\mathbf{v}_{d,0}$ is directed toward the monkey, so that $\mathbf{v}_{m,0} - \mathbf{v}_{d,0} = \mathbf{v}_{d,0}t_0$ .

Subtracting the two equations and using this substitution, we find $\mathbf{x}_m - \mathbf{x}_d = \mathbf{v}_{d,0}(t_0 - t) + \mathbf{v}_{m,0} t.$ The monkey is hit if this expression every becomes zero. Obviously, if the second term $\mathbf{v}_{m,0}$ is not parallel to the vector $\mathbf{v}_{d,0}$ , this will never be the case.

On the other hand, if $\mathbf{v}_{m,0} = \alpha\mathbf{v}_{d,0}$ is parallel to the direction in which the dart is launched, the equation becomes $\mathbf{v}_{d,0}(t_0 + (1-\alpha)t) = \mathbf{0},$ so that the monkey will be hit at time $t = t_0/(1-\alpha)$ . (There is no positive solution, i.e. no hit, if $\alpha \geq 1$ . This means that the monkey launches itself in the same direction as the dart, but at a higher speed!)