For this case, let's
assume that the pot spends exactly half of its time going up, and half going
down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take
the bottom of the window to be zero on a vertical axis pointing upward. All calculations
will be made in reference to this coordinate system. <span>
An initial condition has been supplied by the problem:
s=1.80m when t=0.245s
<span>This means that it takes the pot 0.245 seconds to travel
upward 1.8m. Knowing that the gravitational acceleration acts downward
constantly at 9.81m/s^2, and based on this information we can use the formula:
s=(v)(t)+(1/2)(a)(t^2)
to solve for v, the initial velocity of the pot as it enters
the cat's view through the window. Substituting and solving (note that
gravitational acceleration is negative since this is opposite our coordinate
orientation):
(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2
v=8.549m/s
<span>Now we know the initial velocity of the pot right when it
enters the view of the window. We know that at the apex of its flight, the
pot's velocity will be v=0, and using this piece of information we can use the
kinematic equation:
(v final)=(v initial)+(a)(t)
to solve for the time it will take for the pot to reach the
apex of its flight. Because (v final)=0, this equation will look like
0=(v)+(a)(t)
Substituting and solving for t:
0=(8.549m/s)+(-9.81m/s^2)(t)
t=0.8714s
<span>Using this information and the kinematic equation we can find
the total height of the pot’s flight:
s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>
s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2
s=3.725m<span>
This distance is measured from the bottom of the window, and
so we will need to subtract 1.80m from it to find the distance from the top of
the window:
3.725m – 1.8m=1.925m</span>
Answer:
<span>1.925m</span>
Plans used for work that has to do with construction in or around Earth are called, “Civil Plans.”
Hope this helped!
The best and most correct answer among the choices provided by the question is the second choice. The rope will not be broken until the net force acting on it is not equal to zero anymore. I hope my answer has come to your help. God bless and have a nice day ahead!
It would be D (hoping the picture you posted is what you needed help with
