Asked by john
A stunt woman sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The speed of the horse is 13.5 m/s, and the woman is initially 2.60 m above the level of the saddle.
(a) What must be the horizontal distance between the saddle and limb when the woman makes her move?
m
(b) How long is she in the air?
s
(a) What must be the horizontal distance between the saddle and limb when the woman makes her move?
m
(b) How long is she in the air?
s
Answers
Answered by
Henry
a. h = Vo*t + 0.5g*t^2.= 2.60 m.
0 + 4.9t^2 = 2.60,
t^2 = 0.531,
Tf = 0.728 s = Fall time.
d = V*t = 13.5m/s * 0.728 s. = 9.83 m.
b. Tf = 0.728 s = Fall tm3e.
0 + 4.9t^2 = 2.60,
t^2 = 0.531,
Tf = 0.728 s = Fall time.
d = V*t = 13.5m/s * 0.728 s. = 9.83 m.
b. Tf = 0.728 s = Fall tm3e.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.