Question

ou step off the limb of a tree clinging to a 30-m-long vine that is attached to another limb at the same height and 30-m distant. Assuming air resistance is negligible, how fast are you gaining speed at the instant the vine makes an angle of 25 degrees with the vertical during your descent

Answer 1

v = 7.42 m / s

Step-by-step explanation:

For this exercise we will use the conservation of mechanical energy.

Starting point. Before jumping from the tree

Em₀ = U = m g h

Final point . Part of the trajectory at 25º

Em_{f} = K + U = ½ m v2 + m g y

as they indicate that there is no air resistance, mechanical energy is conserved

Em₀ = Em_{f}

m g h = ½ m v² + m g y

v² = 2g (h - y)

Let's use trigonometry to find the height that has descended, how the angle is measured with respect to the vertical

cos 25 = y / L

y = L cos 25

we substitute

v² = 2 g (h - L cos 25)

for this case h = L = 30 m

v2 = 2g L (1- cos25)

let's calculate

v² = 2 9.8 30 (1 -cos 25)

v = √55.09

v = 7.42 m / s