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 64¡ã above the horizontal. With this launch angle, a skier attains a height of 10 m above the end of the ramp. What is the skier's launch speed?
V sin 64 is the vertical velocity component at launch.
To attain a vertical rise of H = 10 m,
(1/2) M (V sin64)^2 = M g H
H = V^2*sin^2(64)/2g = 0.4039 V^2/g
Solve for V
To find the launch speed of the skier, we can use the principles of projectile motion. Let's break down the problem and use the following variables:
Launch angle (θ) = 64° (given)
Height (h) = 10 m (given)
Launch speed (v₀) = ?
First, let's analyze the vertical motion of the skier. We can use the following equation:
h = (v₀² * sin²θ) / (2 * g),
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the given values:
10 = (v₀² * sin²64°) / (2 * 9.8).
Now, let's solve this equation for v₀.
1. Multiply both sides of the equation by (2 * 9.8):
20 * 9.8 = v₀² * sin²64°.
2. Divide both sides of the equation by sin²64°:
(20 * 9.8) / sin²64° = v₀².
3. Take the square root of both sides to isolate v₀:
v₀ = √((20 * 9.8) / sin²64°).
Now, let's calculate the launch speed using the formula:
v₀ = √((20 * 9.8) / sin²64°).
Using a scientific calculator, we can input the values and calculate the result.
The launch speed of the skier is approximately 20.58 m/s (rounded to two decimal places).
Therefore, the skier's launch speed is approximately 20.58 m/s.