A trapeze artist swings from one 35-ft long trapeze to another trapeze. She releases one trapeze at a 30 degree angle relative to the vertical. She flies through the air, catching the other trapeze 18 ft away at the same elevation. What was her release velocity?

Answer 1

As you know, the equation for the height is

y = -gsec^2(θ)/2v^2 x^2 + x tanθ
In our case, that is

y = -32(4/3)/2v^2 x^2 + 1/√3 x

If we set (0,0) to be the point of release, we want (18,0) to be the point of catch. So,

-32(4/3)/2v^2 (324) + 1/√3 (18) = 0
v = 25.79 ft/s

Answer 2

Thanks!