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 61° above the horizontal. With this launch angle, a skier attains a height of 15 m above the end of the ramp. What is the skier's launch speed?
2007-10-11 14:31:02 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Hi Zella:
First lets calculate the speed upward.
the Energy to go up 15 m is
PE=mgh=m*9.8 m/sec^2* 15 m
=m*147m^2/sec^2
The velocity going up at the end of the ramp is
1/2 m v^2=m*147m^2/sec^2
1/2 v^2=147m^2/sec^2
v=8.57m/sec
The total velocity will be
tv=v/sin (61)
tv=8.57m/sec*1.14
=9.77 m.sec
2007-10-11 14:52:54 · answer #1 · answered by Frst Grade Rocks! Ω 7 · 1⤊ 0⤋