An archer pulls her bowstring back 0.396 m by exerting a force that increases uniformly from zero to 237 N. (a) What is the equi
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Answer:
Vf = 69.61 m/s
Explanation:
We will use the third equation of motion to solve this problem:
where,
g = acceleration due to gravity = 9.81 m/s²
h = height of cliff = 247 m
Vf = final velocity = ?
Vi = initial velocity = 0 m/s (boulder breaks loose from rest)
Therefore,
<u>Vf = 69.61 m/s</u>
assuming north-south along the Y-axis and east-west along the X-axis
X = total X-displacement
from the graph , total displacement along X-direction is given as
X = 0 - 20 + 60 Cos45 + 0
X = 42.42 - 20
X = 22.42 m
Y = total Y- displacement
from the graph , total displacement along Y-direction is given as
Y = 40 + 0 + 60 Sin45 + 50
Y = 90 + 42.42
Y = 132.42 m
magnitude of the net displacement vector using Pythagorean theorem is given as
magnitude : Sqrt(X² + Y²) = Sqrt(22.42² + 132.42²) = 134.31 m
Direction : tan⁻¹(Y/X) = tan⁻¹(132.42/22.42) = 80.4 deg north of east
