Answer:
Approximately [tex]110\; {\rm m\cdot s^{-1}}[/tex], assuming that the time required for the person to achieve the given velocity is negligible.
Explanation:
Under the assumptions, momentum should be conserved right after the jump. In other words, the momentum of the person and the chair, combined, should be the same before and after the jump.
Let [tex]m_{1}[/tex] denote the mass of the person and let [tex]m_{2}[/tex] denote the mass of the chair. Let [tex]u_{1}[/tex] and [tex]u_{2}[/tex] denote the velocity of the person and the chair before the jump, and let [tex]v_{1}[/tex] and [tex]v_{2}[/tex] denote the velocity after the jump, respectively.
When an object of mass [tex]m[/tex] travels at a velocity of [tex]v[/tex], the momentum of that object would be [tex]p = m\, v[/tex].
Right before the jump, the momentum of the person and the chair, combined, would be:
[tex]m_{1}\, u_{1} + m_{2}\, u_{2}[/tex].
Right after the jump, this value would be:
[tex]m_{1}\, v_{1} + m_{2}\, v_{2}[/tex].
Since momentum is conserved under the assumptions:
[tex]m_{1} \, v_{1} + m_{2}\, v_{2} = m_{1}\, u_{1} + m_{2}\, u_{2}[/tex].
However, it is also given that the person and the chair were stationary before the jump. In other words, [tex]u_{1} = u_{2} = 0[/tex]. Hence:
[tex]m_{1} \, v_{1} + m_{2}\, v_{2} = m_{1}\, u_{1} + m_{2}\, u_{2} = 0[/tex].
[tex]m_{1} \, v_{1} + m_{2}\, v_{2} = 0[/tex],
Rearrange this equation to express [tex]v_{2}[/tex] (velocity of the chair right after the jump) in terms of [tex]m_{1}[/tex], [tex]v_{1}[/tex], and [tex]m_{2}[/tex]:
[tex]\begin{aligned}v_{2} &= \frac{-m_{1}\, v_{1}}{m_{2}} \\ &= \frac{-(34\; {\rm kg})\, (8.1\; {\rm m\cdot s^{-1}})}{(2.5\; {\rm kg})} \\ &\approx (-110)\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
(The velocity of the person is positive. The fact that the velocity of the chair would be negative means that the motion of the chair would be in the opposite direction of the motion of the person.)
In other words, the velocity of the chair right after the jump would be approximately [tex]110\; {\rm m\cdot s^{-1}}[/tex] opposite to the direction of motion of the person.
