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simple velocity/acceleration question?
a branch located 90m up a tree falls from rest. what is its speed when it hits the ground
i am interested in the work/equation. best work gets chosen
2 Answers
- DavidK93Lv 710 years agoFavorite Answer
The best equation to use here is v^2 = (v_0)^2 + 2ax, where v is final velocity, v_0 is initial velocity, a is acceleration, and x is distance traveled. In this case, v_0 = 0 because the branch falls from rest, a = g = 9.8 m/s^2, and x = h = 90 m.
v^2 = (v_0)^2 + 2ax
v^2 = 0^2 + 2gh
v^2 = 2gh
v = sqrt(2gh) = sqrt(2*9.8*90) = sqrt(1764) = 42 m/s
- Anonymous10 years ago
V(final)^2 - V(initial)^2 = 2 x acceleration x distance
V(final)^2 = (2)(9.81)(90)
V(final) = 42.02 m/s
Source(s): my brain