A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en
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Answer:
3.467 s
Explanation:
given,
distance , d = 49 mm = 0.049 m
initial speed of the of the rock, v = 17 m/s
time taken by the Heather rock to reach water
using equation of motion
taking downward as negative
4.9 t² + 17 t - 0.049 = 0
now,
t₁ = -3.47 s , 0.0028 s
rejecting negative values
t₁ = 0.0028 s
now, time taken by the ball of Jerry
using equation of motion
taking downward as negative
4.9 t² - 17 t - 0.049 = 0
now,
t₂ = 3.47 s ,-0.0028 s
rejecting negative values
t₂ = 3.47 s
now, time elapsed is = t₂ - t₁ = 3.47 - 0.0028 = 3.467 s
In triangle ABC , using Pythagorean theorem
BC = sqrt(AB² + AC²)
r = sqrt(y² + x²) eq-1
taking derivative both side relative to "t"
dr/dt = (1/(2 sqrt(y² + x²) ) ) (2 y (dy/dt) + 2 x (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) (2 (0.5) (dy/dt) + 2 (0.5) (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) ( v₁ + v₂)
15= (1/(2 sqrt(0.5² + 0.5²) ) ) ( - 30 + v₂)
v₂ = 51.2 m/s
