Respuesta :
Answer:
Explanation:
Given
length of window [tex]h=2.9\ m[/tex]
time Frame for which rock can be seen is [tex]\Delta t=0.134\ s[/tex]
Suppose h is height above which rock is dropped
Time taken to cover [tex]h+2.9 is t_1[/tex]
so using equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
where y=displacement
u=initial velocity
a=acceleration
t=time
time taken to travel h is
[tex]h=0+0.5\times g\times (t_2)^2---2[/tex]
Subtract 1 and 2 we get
[tex]2.9=0.5g(t_1^2-t_2^2)[/tex]
[tex]5.8=g(t_1+t_2)(t_1-t_2))[/tex]
and from equation [tex]t_1-t_2=0.134\ s[/tex]
so [tex]t_1+t_2=\frac{5.8}{9.8\times 0.134}[/tex]
[tex]t_1+t_2=4.416\ s[/tex]
and [tex]t_1=t_2+\Delta t[/tex]
so [tex]t_2+\Delta t+t_2=4.416[/tex]
[tex]2t_2+0.134=4.416[/tex]
[tex]t_2=0.5\times 4.282[/tex]
[tex]t_2=2.141\ s[/tex]
substitute the value of [tex]t_2[/tex] in equation 2
[tex]h=0.5\times 9.8\times (2.141)^2[/tex]
[tex]h=22.46\ m[/tex]
The rock was dropped from a height of 23.69m from the top of the window.
Let the initial velocity of the rock at the top of the window be [tex]v[/tex].
Given
the height of the window [tex]h=2.9m[/tex] and the time taken to pass the window is [tex]t=0.134s[/tex]
Kinematics equation of motion:
so from the second equation of motion:
[tex]h=vt+\frac{1}{2}gt^2[/tex]
[tex]2.9=0.134v+0.5\times9.8\times(0.134)^2\\\\v=21.55m/s[/tex]
Now when the rock 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\\\\(21.55)^2=2\times9.8\times d\\\\d=23.69m[/tex]is the height above the top of the window from which the rock falls.
Find out more about equation of motion:
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