Answer:
the distance of the top of the window below the windowsill from which the flowerpot fell is 0.31 m
Explanation:
given information:
time, t = 0.420 s
height, h = 1.9 m
the the vertical motion equation:
h = [tex]v_{0}[/tex]t + [tex]\frac{1}{2} gt^{2}[/tex]
where
h = height (m)
[tex]v_{0}[/tex] = initial velocity (m/s)
t = time (s)
g = gravitational force (9.8 [tex]m/s^{2}[/tex])
first we calculate the velocity of the flowers pot when it reaches top window
h = [tex]v_{0}[/tex]t + [tex]\frac{1}{2} gt^{2}[/tex]
1.9 = [tex]v_{0}[/tex](0.42) + ([tex]\frac{1}{2} 9.8 0.42^{2}[/tex])
[tex]v_{0}[/tex] = (1.9 - ([tex]\frac{1}{2} 9.8 0.42^{2}[/tex]))/0.42
= 2.47 m/s
now we can find the distance of the windowsil
[tex]v^{2} =v_{0} ^{2}[/tex] + 2gh
v = 0, thus
2gh = - [tex]v_{0}^{2}[/tex]
h = -[tex]v_{0}^{2}[/tex]/2g
= [tex](- 2.47^{2} )[/tex] / (2 x 9.8)
= 0.31 m
The top of the window is 0.308m below the windowsill from which the flowerpot fell.
Let the initial velocity of the flowerpot at the top of the window be [tex]v[/tex].
Given that the height of the window [tex]h=1.9m[/tex] and the time taken to pass the window is [tex]t=0.42s[/tex]
so from the second equation of motion:
[tex]h=vt+\frac{1}{2}gt^2[/tex]
[tex]1.9=0.42v+0.5\times9.8\times(0.42)^2\\\\v=2.46m/s[/tex]
Now when the flowerpot falls from its original position, its initial velocity is [tex]u=0[/tex], let the distance between the position from where it falls and the top of the window be [tex]d[/tex],
so, applying the third equation of motion we get that:
[tex]v^2=u^2+2gd\\\\(2.46)^2=2\times9.8\times d\\\\d=0.308m[/tex]is the height above the top of the window from which the flowerpot falls.
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