A circus performer begins his act by walking out along a nearly horizontal high wire. He slips and falls to the safety net, 25.0 ft below. The magnitude of his displacement vector from beginning of the walk to the net is 26.7 ft. How far out along the high wire did he walk?
D = X^2 + Y^2 = 26.7^2
X^2 + 25^2 = 26.7^2
Solve for X.
To determine how far out along the high wire the performer walked, we need to calculate the horizontal displacement.
Given:
Vertical displacement (fall to safety net) = 25.0 ft
Magnitude of displacement vector (from beginning to net) = 26.7 ft
Using the Pythagorean theorem, we can calculate the horizontal displacement:
horizontal displacement^2 + vertical displacement^2 = magnitude of displacement vector^2
Let's solve for the horizontal displacement:
horizontal displacement^2 + 25.0^2 = 26.7^2
horizontal displacement^2 + 625 = 712.89
horizontal displacement^2 = 712.89 - 625
horizontal displacement^2 = 87.89
Taking the square root of both sides:
horizontal displacement = sqrt(87.89)
horizontal displacement ≈ 9.37 ft
Therefore, the performer walked approximately 9.37 ft out along the high wire.
To find out how far the circus performer walked out along the high wire, we need to calculate the magnitude of his displacement vector. The magnitude of the displacement vector is the distance between the starting point and the ending point of the motion.
Given:
- Displacement vector magnitude = 26.7 ft
- Height of the safety net = 25.0 ft
We can use the Pythagorean theorem to solve for the distance the performer walked out along the high wire. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Let's represent the distance walked along the high wire as "x" and the height of the safety net as "y". According to the problem, y = 25.0 ft and the magnitude of the displacement vector is 26.7 ft. Using the Pythagorean theorem, we have:
x^2 + y^2 = (displacement vector magnitude)^2
Substituting the given values:
x^2 + (25.0 ft)^2 = (26.7 ft)^2
x^2 + 625.0 ft^2 = 712.89 ft^2
x^2 = 712.89 ft^2 - 625.0 ft^2
x^2 = 87.89 ft^2
Taking the square root of both sides to solve for x:
x = √(87.89 ft^2)
x ≈ 9.37 ft
Therefore, the circus performer walked out approximately 9.37 ft along the high wire before slipping and falling to the safety net.