Respuesta :
This probably should've been posted in the Physics section, but I'll help anyway.
There are several kinematic equations:
V = Vo + at
Δx = (Vo)t + (1/2)at^2
There are a bit more than that, but here we go!
We'll focus on this one first, but keep in mind the variables are the same!:
V = Vo + at
Where,
Δx = The Change in Position (Distance)
Vo = The Initial Velocity
t = time
a = acceleration (gravity)
Let's plug in our values!
Δx = 20 - 0 = 20m (0 is our final distance from Earth. AKA when we stop falling)
Vo = 0 m/s(The words, "From rest" tell us this)
t = ?
a = 9.8 m/s^2 (9.8 m/s^2 is just our acceleration due to the gravity of Earth)
The question asks when is the speed of the object 9.8 meters per second?
Plug in 9.8 m/s as our final velocity, 0 as our initial veloicty, and 9.8 as our acceleration. Solve for t.
9.8 m/s = 0 + 9.8*t
t = (9.8 m/s) / (9.8 m/s^2)
t = 1s
So our answer is:
2.
Close one between 1 and 2, but 2 is the correct answer, because 1 would mean that during every moment of time from 0s - 1s our speed would have to be 9.8m/s. Which is obviously not true, because it started from REST.
3. Is just wrong, because that would mean the acceleration is constant, which it literally cannot be if it is falling.
4. Can be found to be wrong if you plug it into the problem as Δx (to be exact you would plug in (20 - 9.8) and then solve for time. From there you would plug time into the other equation and fine velocity. Although it's pretty obvious that if you were accelerating at 9.8 m/s^2 if you only gained velocity from your acceleration due to gravity (which in this case you do, because we start from rest) then you'd realize that after 1 second you would be at 9.8 m/s. :)
There are several kinematic equations:
V = Vo + at
Δx = (Vo)t + (1/2)at^2
There are a bit more than that, but here we go!
We'll focus on this one first, but keep in mind the variables are the same!:
V = Vo + at
Where,
Δx = The Change in Position (Distance)
Vo = The Initial Velocity
t = time
a = acceleration (gravity)
Let's plug in our values!
Δx = 20 - 0 = 20m (0 is our final distance from Earth. AKA when we stop falling)
Vo = 0 m/s(The words, "From rest" tell us this)
t = ?
a = 9.8 m/s^2 (9.8 m/s^2 is just our acceleration due to the gravity of Earth)
The question asks when is the speed of the object 9.8 meters per second?
Plug in 9.8 m/s as our final velocity, 0 as our initial veloicty, and 9.8 as our acceleration. Solve for t.
9.8 m/s = 0 + 9.8*t
t = (9.8 m/s) / (9.8 m/s^2)
t = 1s
So our answer is:
2.
Close one between 1 and 2, but 2 is the correct answer, because 1 would mean that during every moment of time from 0s - 1s our speed would have to be 9.8m/s. Which is obviously not true, because it started from REST.
3. Is just wrong, because that would mean the acceleration is constant, which it literally cannot be if it is falling.
4. Can be found to be wrong if you plug it into the problem as Δx (to be exact you would plug in (20 - 9.8) and then solve for time. From there you would plug time into the other equation and fine velocity. Although it's pretty obvious that if you were accelerating at 9.8 m/s^2 if you only gained velocity from your acceleration due to gravity (which in this case you do, because we start from rest) then you'd realize that after 1 second you would be at 9.8 m/s. :)
Answer:
The correct option is 2. at the end of its first second of fall.
Step-by-step explanation:
Consider the provided information:
The equation of motion is:
[tex]V_{final}=V_{initial}+at[/tex]
Object is dropped from the rest. Therefore [tex]V_{initial} = 0\ m/s[/tex].
a is 9.8 meters per second because a is acceleration due to gravity.
We need to find the time at which the speed of object is 9.8 meters per second. Thus [tex]V_{final}= 9.8\ m/s[/tex]
Thus, substitute the respective values in the above equation.
[tex]9.8 \ m/s =0 +(9.8 \ m/s^{2}})t[/tex]
[tex]1s=t[/tex]
This means that, if the object has a velocity of 0 meter per second now, it will have the velocity of 9.8 meter per second at the end of its first second.
Hence, the correct option is 2. at the end of its first second of fall.
