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 63° above the horizontal. With this launch angle, a skier attains a height of 13.6 m above the end of the ramp. What is the skier's launch speed?
let s = speed at launch
v = 0 at top = s sin 63 - g t
so at top
t = s sin 63/g = .0909 s
h = 13.6 = s sin 63 t - 4.9 t^2
13.6 = .081s^2 - .0405 s^2
s^2 = 336
s = 18.3 m/s
To find the skier's launch speed, we can use the projectile motion equations. We'll start by finding the initial vertical velocity (v₀y).
Since the launch ramp is directed 63° above the horizontal, we can decompose the velocity into horizontal (v₀x) and vertical (v₀y) components. The initial velocity (v₀) can be written as:
v₀x = v₀ * cos(θ)
v₀y = v₀ * sin(θ)
Where:
v₀x is the initial horizontal velocity
v₀y is the initial vertical velocity
v₀ is the initial velocity that we're trying to find
θ is the launch angle which is 63°
Next, we'll use the equation for vertical displacement to find the initial vertical velocity (v₀y):
Δy = v₀y * t + (1/2) * g * t²
Where:
Δy is the vertical displacement (13.6 m in this case)
t is the time of flight
g is the acceleration due to gravity (9.8 m/s²)
Since the skier's vertical displacement is at the maximum height, the final vertical velocity (vfy) is 0 m/s.
0 = v₀y - g * t
Now, we can substitute v₀y in terms of v₀ and θ:
0 = v₀ * sin(θ) - g * t
We also know that the horizontal and vertical components of velocity are related by time:
v₀x = v₀ * cos(θ)
v₀y = v₀ * sin(θ)
Rearranging both equations, we can solve for t:
t = v₀x / (v₀ * cos(θ))
Now, we can substitute the expression for t into the equation for vertical displacement:
0 = v₀ * sin(θ) - g * (v₀x / (v₀ * cos(θ)))
Simplifying the equation:
0 = sin(θ) - g * (v₀x / v₀)
Now, we need to solve for v₀, the initial velocity:
v₀ = v₀x / sin(θ)
Finally, substituting v₀x = v₀ * cos(θ) into the equation:
v₀ = (v₀ * cos(θ)) / sin(θ)
Now we can solve this equation to find the skier's launch speed.
To determine the skier's launch speed, we can use the principles of projectile motion. The key information we need is the launch angle and the maximum height reached by the skier.
Let's break down the problem and find a systematic approach to solve it:
1. Define the given data:
- Launch angle (θ) = 63°
- Maximum height (h) = 13.6 m
- Acceleration due to gravity (g) = 9.8 m/s²
2. We can use the kinematic equations of projectile motion to solve for the launch speed (v₀). The two relevant equations are:
- Vertical height: h = (v₀² * sin²θ) / (2 * g)
- Vertical velocity: v_y = v₀ * sinθ
3. Rearrange the first equation to solve for v₀:
- v₀² = (2 * g * h) / sin²θ
- v₀ = √[(2 * g * h) / sin²θ]
4. Plug in the values into the equation:
- v₀ = √[(2 * 9.8 * 13.6) / sin²63°]
- v₀ ≈ 18.2 m/s
Therefore, the skier's launch speed is approximately 18.2 m/s.