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 degree above the horizontal. With this launch angle, a skier attains a height of 11.4 m above the end of the ramp. What is the skier's launch speed?

Answer 1

To find the skier's launch speed in the aerials competition, we can use projectile motion equations. By using the given values of the launch angle and the height reached, we can solve for the initial launch speed of the skier.

Step-by-step explanation:

To find the skier's launch speed, we can use the projectile motion equations. The height reached by the skier is the vertical component of the launch, and the launch angle is given as 63 degrees above the horizontal. We can use the equation:

h = (V02sin2θ0) / (2g)

where h is the height reached, V0 is the initial launch speed, θ0 is the launch angle, and g is the acceleration due to gravity.

Plugging in the given values, we have:

11.4 m = (V02sin263) / (2 * 9.8 m/s2)

Simplifying the equation and solving for V0, we find:

V0 = sqrt((11.4 m * 2 * 9.8 m/s2) / sin263)

Using a calculator, we can calculate V0 to be approximately 23.4 m/s.