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 66° 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?

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 66° 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

The 12 m goes into Potential energy, whilst the horizontal velocity is constant.

1/2 m V^2=mg(12)+1/2 m Vcos66

solve for V.

V^2-Vcos66-24*9.8=0

Use the quadratic equation

Answer 2

the skier's upward velocity is v, so the upward component is v*sin 66° = .9135v

so, if the ramp ends at (0,0), the skier's height is

h = .9135v*t - 4.9t^2
= t(.9135v-4.9t)
roots are at t=0,0.1864v

max h is achieved midway between the roots, at t = 0.0932v

12 = .9135v(.0932v) - 4.9(.0932v)^2
v = 16.7884

Answer 3

To find the skier's launch speed, we can use some basic principles of projectile motion. We need to determine the initial vertical velocity component and the initial horizontal velocity component.

Let's start by resolving the given launch angle into its horizontal and vertical components. The launch angle of 66° is measured with respect to the horizontal direction. The vertical component can be calculated using trigonometry:

Vertical component velocity (Vy) = Launch speed (V) * sin(θ)

Where θ is the launch angle in radians.

Now that we have the vertical component velocity, we can find the time it takes for the skier to reach the maximum height (12 m) using the kinematic equation:

Final velocity (Vyf) = Initial velocity (Vyi) + (acceleration (ay) * time (t))

At the maximum height, the final velocity will be zero since the skier momentarily stops moving vertically. The acceleration (ay) is equal to the acceleration due to gravity (-9.8 m/s^2).

By substituting the given values, we can solve for the time (t).

0 = Vy - (9.8 m/s^2 * t)

On rearranging the equation:

t = Vy / 9.8 m/s^2

Next, we can use the time (t) to find the horizontal component velocity (Vx) using the kinematic equation:

Horizontal distance (dx) = Initial velocity (Vxi) * time (t) + (acceleration (ax) * time (t)^2 / 2)

Since the skier is not accelerating horizontally (ax = 0), the equation becomes:

dx = Vx * t

Rearranging for Vx:

Vx = dx / t

Now we need to determine the horizontal distance (dx) traveled. It is not explicitly given in the question, but it is related to the height (12 m) by the launch angle.

Using trigonometry again, we can find:

Horizontal distance (dx) = Height (h) / tan(θ)

Now we have the horizontal component velocity (Vx) and vertical component velocity (Vy), we can find the total launch speed (V) using the Pythagorean theorem:

V = sqrt(Vx^2 + Vy^2)

Substituting the values, you can calculate the skier's launch speed.