Answer to Question #157922 in Physics for Titilayo

Question

A ski jumper takes off from a horizontal ramp 5 meters in the air. The ramp is directly next to a steep drop of 70 meters. The skier lands 146 meters away from the base of the ramp. What was the skier’s original horizontal velocity?

Answer 1

$$h=0.5gt^2\\t=\sqrt{\frac{2h}{g}}\\x=vt=v\sqrt{\frac{2h}{g}} \\146=v\sqrt{\frac{2(70+5)}{9.8}}\\v=37.3\frac{m}{s}$$