Answer
He should jump when the truck is approximately 1.5 poles away.
Work Step by Step
Agent Bond's initial position is considered to be $y=0$.
We choose downward to be the positive direction. The initial velocity of Agent Bond is $v=0$ while his acceleration is $a=g$. His displacement if he jumped onto the truck would be:
$y=15m-3.5m=11.5m$
Duration of Agent Bond's fall can be found by replacing $x$ with $y$ in the following formula:
$$y=y_{0}+v_{0}t+\frac{1}{2}at^{2}$$
$$t=\sqrt (\frac{2y}{a})=\sqrt (\frac{2(11.5m)}{9.8m/s^{2}})=1.532s$$
Agent bond needs to jump when the truck is $d$ distance away. The truck is approaching at a constant velocity of $25m/s^2$. So we find $d$:
$$d=vt=(25m/s^2)(1.532s)=38.3m$$
Converting this distance into ''poles'':
$$(38.3m)(\frac{1pole}{25m})=1.532 poles\approx1.5 poles$$
So, Agent Bond should jump when the truck is approximately 1.5 poles away.
