does anyone kno physics?

Question

The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply upward at the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. In the women's competition, the end of a typical launch ramp is directed 58° above the horizontal. With this launch angle, a skier attains a height of 12 m above the end of the ramp. What is the skier's launch speed?

Answer 1

Vh = Vsin58°
Vh = √(2gh)
V = (√(2gh))/sin58°
V = (√(2*9.80665*12))/0.8480481
V ≈ 18.090 m/s