## Fundamentals of Physics Extended (10th Edition)

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