Respuesta :
Answer:
Explanation:
Given two objects are dropped simultaneously
Object A is 10 m higher than object B therefore
Distance covered by object A is given by
[tex]y_a(t)[/tex] is given by
[tex]y=ut+\frac{1}{2}at^2[/tex]
where y=displacement
u=initial velocity
a=acceleration
t=time
[tex]y_a(t)=0+0.5gt^2--1[/tex]
for object B
[tex]y_b(t)=0+0.5gt^2--2[/tex]
Subtract 1 and 2 we get
[tex] y_a(t)-y_b(t)=0[/tex]
i.e. they will travel equal distance in equal time and distance between them remain 10 m until object B hits the ground
The distance between the object remains constant during descent until object B hits the ground.
Equations of motion:
Let us use the equations of motion to calculate the equation for the height of the object under gravity.
Object A, dropped from the roof, let the height of the roof be H.
The initial speed of A will be zero, u = 0,
then from the second equation of motion, we get that the distance traveled by the object A in time t is :
[tex]x = ut + \frac{1}{2} gt^2\\\\x=\frac{1}{2}gt^2[/tex]
so the final height of A after time t is:
d = H-x
[tex]d = H-\frac{1}{2}gt^2[/tex]
Now, object B is dropped from a height (H -10) m. Similarly, like A, the distance traveled by the object B in time t is:
[tex]x=\frac{1}{2}gt^2[/tex]
so the final height of B after time t is:
d' = H - 10 - x
d' = [tex]d'=H-10-\frac{1}{2} gt^2[/tex]
Now, we get that:
d - d' = 10m
So the difference between heights remains constant at 10m.
Object B hits the ground first as it is closer to the ground, and obviously, after that, the difference between height will keep decreasing until it becomes zero when both are on the ground.
Learn more about equations of motion:
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