An aerialist on a high platform holds on to a trapeze
attached to a support by an 8.00-m cord. (See the drawing.)
Just before he jumps off the platform, the cord makes an
angle of 41.0° degrees with the vertical. He jumps, swings
down, then back up, releasing the trapeze at the instant it is
0.750 m below its initial height. Calculate the angle θ that
the trapeze cord makes with the vertical at this instant.
At the start, he is d1 = 8 cos41° below the platform.
On the release, he is d2 = 8 cos41° + 0.75 below the platform
So cosθ = d2/8
Now finish it off
To calculate the angle θ that the trapeze cord makes with the vertical at the instant it is 0.750 m below its initial height, we can use trigonometry.
Let's define the variables:
θ: angle the trapeze cord makes with the vertical at the instant it is 0.750 m below its initial height
L: length of the trapeze cord (8.00 m)
h: initial height of the aerialist when he holds onto the trapeze
d: distance below the initial height the trapeze falls before releasing it (0.750 m)
First, we need to calculate the initial height of the aerialist, h:
h = L * cos(θ_initial)
where θ_initial is the initial angle the cord makes with the vertical (41.0°).
h = 8.00 m * cos(41.0°)
h ≈ 5.76 m
Now, we can calculate the vertical distance the trapeze falls, y:
y = h - d
y = 5.76 m - 0.750 m
y ≈ 5.01 m
Next, we'll use the right triangle formed by the vertical distance the trapeze falls and the length of the trapeze cord to calculate the angle θ:
sin(θ) = y / L
θ = arcsin(y / L)
θ = arcsin(5.01 m / 8.00 m)
Using a calculator, we find that:
θ ≈ 37.28°
So, the angle θ that the trapeze cord makes with the vertical at the instant it is 0.750 m below its initial height is approximately 37.28°.