Answer
The halteres would add a distance of $~~55~cm~~$ to the range.
Work Step by Step
We can use conservation of momentum to find the jumper's horizontal speed after releasing the two halteres:
$p_f = p_i$
$(78~kg)(v_{x})+(11~kg)(0) = (89~kg)(9.5~m/s)$
$v_{x} = \frac{(89~kg)(9.5~m/s)}{78~kg}$
$v_x = 10.84~m/s~m/s$
We can find the time it takes the jumper to reach maximum height:
$v_{yf} = v_{y0}+a_y~t$
$t = \frac{v_{yf} - v_{y0}}{a_y}$
$t = \frac{0 - 4.0~m/s}{-9.8~m/s^2}$
$t = 0.408~s$
Note that it will take the jumper this same time to return to the ground.
We can find the jumper's range without using the halteres:
$x = (9.5~m/s)(0.408~s)+(9.5~m/s)(0.408~s)$
$x = 7.75~m$
We can find the jumper's range when using the halteres:
$x = (9.5~m/s)(0.408~s)+(10.84~m/s)(0.408~s)$
$x = 8.30~m$
We can find the distance that the halteres add to the range:
$\Delta x = 8.30~m-7.75~m = 0.55~m = 55~cm$
The halteres would add a distance of $~~55~cm~~$ to the range.
