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
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.
1 answer