Answer
The hang glider's new speed is 15.8 m/s
Work Step by Step
Let $h_1$ be the original height. Let $h_2$ be the lower height. We can use conservation of energy to find the new speed $v_2$:
$mgh_2 + \frac{1}{2}mv_2^2 = mgh_1 + \frac{1}{2}mv_1^2$
$\frac{1}{2}mv_2^2 = mg~(h_1-h_2) + \frac{1}{2}mv_1^2$
$v_2^2 = 2g~(h_1-h_2) + v_1^2$
$v_2 = \sqrt{2g~(h_1-h_2) + v_1^2}$
$v_2 = \sqrt{(2)(9.80~m/s^2)~(8.2~m) + (9.5)^2}$
$v_2 = 15.8~m/s$
The hang glider's new speed is 15.8 m/s.
