A length of rope is stretched between the top edge of a building and a stake in the ground. The rope also touches a tree halfway
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Answer:
a. v(t)= -6.78 + 16.33 b. 16.33 m/s
Step-by-step explanation:
The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)= =
. We now multiply both sides of the equation by the integrating factor.
μv' + μkv = μg ⇒ v' + k
v = g
⇒ [v
]' = g
. Integrating, we have
∫ [v]' = ∫g
v =
+ c
v(t)= + c
.
From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have
9.55 = 9.8 × 15/9 + c = 16.33 + c
c = 9.55 -16.33 = -6.78.
So, v(t)= 16.33 - 6.78. m/s = - 6.78
+ 16.33 m/s
b. Velocity of object at time t = 0.5
At t = 0.5, v = - 6.78 + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s