Question
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?
Answers
Steve
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
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
Joe
Thanks!