the answer for this question is c
The tension in the rope B is determined as 10.9 N.
<h3>
Vertical angle of cable B</h3>
tanθ = (6 - 4)/(5 - 0)
tan θ = (2)/(5)
tan θ = 0.4
θ = arc tan(0.4) = 21.8 ⁰
<h3>Angle between B and C</h3>
θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰
Apply cosine rule to determine the tension in rope B;
A² = B² + C² - 2BC(cos A)
B = C
A² = B² + B² - (2B²)(cos A)
A² = 2B² - 2B²(cos 43.6)
A² = 0.55B²
B² = A²/0.55
B² = 65.3/0.55
B² = 118.73
B = √(118.73)
B = 10.9 N
Thus, the tension in the rope B is determined as 10.9 N.
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Answer:
a) V ≈ 125 m/s; b) Δt = 13.24 s; c) ΔS ≈ 1450 m
Explanation:
a) We have just to calculate the vector resultant.
V² = 106² + 66.2²
V² = 15618.44
V ≈ 125 m/s
b) The time of flight is equal to the time to reach the maximum height summed to the time to reach the land.
In vertical:
V = V₀ + a * t
V = 66.2 - g * t
0 = 66.2 - 9.8 * t
t ≈ 6.76 s
So: Δt = 13.24 s
c) In horizontal:
V = ΔS / Δt
106 = ΔS / 13.52 ⇒ ΔS = 106 * 13.52
ΔS = 106 * 13.52
ΔS = 1433,12
ΔS ≈ 1450 m