#### Answer

43 m/s

#### Work Step by Step

We are given the horizontal range of the jump was 77m. We are also told that the angle to the horizontal is $12^{\circ}$
We use the equation for range:
$R = \frac{v_{0}^2}{g}sin2\theta_{0} $
This becomes
$77 = \frac{v_{0}^2}{9.80}sin(2(12^{\circ})) $ and solving for $v_{0}$ we find that it is 43 m/s