Answer:
about 602 milliseconds
Explanation:
The motion can be approximated by the equation ...
y = -4.9t^2 -22.8t +15.5
where t is the time since the arrow was released, and y is the distance above the ground.
When y=0, the arrow has hit the ground.
Using the quadratic formula, we find ...
t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))
= (22.8 ± √823.64)/(-9.8)
The positive solution is ...
t ≈ 0.60195193
It takes about 602 milliseconds for the arrow to reach the ground.
Answer:
-30=5(x+1) is -7
Explanation:
distribute flip subtract 5 from both sides divide both sides by 5
Answer:
Explanation:
When the craft was stationary , weight will be balanced by tension
T = mg
T = 7000 N
A)
when the craft was being lowered to the seafloor
drag force will act in upper direction , so
T₁ + 1800 = mg
T₁ = mg - 1800
= 7000 - 1800
= 5200 N
52 X 10² N
B)
when the craft was being raised from the seafloor , Tension will act in downward direction
T₂ = mg+ 1800
T₂ = 7000 - 1800
= 8800N
