Question

In the aerials competition in skiing, the competitors 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. The end of a launch ramp is directed 63° above the horizontal. With this launch angle, a skier attains a height of 10.9m above the end of the ramp. What is the skier’s launch speed?

Answer 1

u = 11.6 m/s

Step-by-step explanation:

The end of a launch ramp is directed 63° above the horizontal. A skier attains a height of 10.9 m above the end of the ramp.

Maximum height, H = 10.9

Let v is the launch speed of the skier. The maximum height attained by the projectile is given by :

`H=(u^2\ sin^2\theta)/(g)`

`10.9=(u^2\ sin^2(63))/(9.8)`

u = 11.6 m/s

So, the launch speed of the skier is 11.6 m/s. Hence, this is the required solution.