vf = vi + at
vf – vi = at<span>
<span>vi= 0, vf=26 and afor nil = 9.8m/s2</span></span>
26 = 9.8t
t =<span> 26 / 9.8 = 2.65 s
Now we know the total time, so we can calculate the time 1
second before it hit the ground.
<span>= 2.65 -1 = 1.65s
<span>Now again using the same equation, vf = vi+at, we can find vf
vi = 0, a = 9.8 t=1.65</span></span></span>
vf = 0 + 9.8(1.65) =
16.17 m/s<span>
</span><span>So,
the nail is traveling with the speed of 16.17m/s 1 second before it hits the
ground.</span>
Answer:
The x-coordinate of the particle is 24 m.
Explanation:
In order to obtain the x-coordinate of the particle, you have to apply the equations for Two Dimension Motion
Xf=Xo+Voxt+0.5axt²(I)
Yf=Yo+Voyt+0.5ayt² (II)
Where Xo, Yo are the initial positions, Xf and Yf are the final positions, Vox and Voy are the initial velocities, ax and ay are the accerelations in x and y directions, t is the time.
The particle starts from rest from the origin, therefore:
Vox=Voy=0
Xo=Yo=0
Replacing Yf=12, Yo=0 and Voy=0 in (I) and solving for t:
12=0+(0)t+ 0.5(1.0)t²
12=0.5t²
Dividing by 0.5 and extracting thr squareroot both sides:
t=√12/0.5
t=√24 = 2√6
Replacing t=2√6, ax=2.0,Xo=0 and Vox=0 in (I) to obain the x-coordinate:
Xf=0+0t+0.5(2.0)(2√6)²
Xf= 24 m
