The solution would be like
this for this specific problem:
<span>v = ? </span><span>
<span>u = 0.0 m/s </span>
<span>a = 9.8 m/s^2 </span>
<span>s = 56.1 m </span></span>
<span>v^2 = (0.0 m/s)^2 + [2 *
(9.8 m/s^2) * (56 m) ] </span><span>
<span>v^2 = 2 * (9.8 m/s^2) * (56 m) </span>
<span>v^2 = 1,097.6 m^2/s^2 </span>
<span>v = SQRT {1,097.6 m^2/s^2 } </span></span>
v = 33.1 m/s
<span>v = u + at </span>
<span>(v - u) / a = t </span>
[ (33.1 m/s) - (0.0 m/s)
] / (9.8 m/s^2) = 3.38 seconds
If the pigeon is 56.0 m below the initial position of the
falcon, it will take 3.38 seconds for the falcon to reach the pigeon. I am
hoping that this answer has satisfied your query and it will be able to help
you in your endeavor, and if you would like, feel free to ask another question.
In my personal opinion, avoiding in not an effective conflict management strategy. When you are avoiding a conflict, you are really just pushing it to the side and it will need to be dealt with sooner or later. Avoiding a conflict is also ammeter thing to do as well, you should take responsibility and handle any situation.
Answer:
Explanation:
Given
car A had a head start of 
and it starts at x=0 and t=0
Car B has to travel a distance of 
where
is the distance travel by car A in time t
distance travel by car A is

For car B with speed 



Time taken by the water balloon to reach the bottom will be given as

here we know that


now by the above formula



now in the same time interval we can say the distance moved by it will be


so it will fall at a distance 15.7 m from its initial position
Answer:
mass of box 1 = 2.20 kg
mass of box 2 = 5.93 kg
Explanation:
Let the mass of box 1 and box 2 is respectively
and 
so we will have
Force applied on box 1 then acceleration



Now we know that contact force between them in above case is given as



now we have
