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September 10, 2009 02:11 AM
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Ok .. for this we are using the Equations of Motion to solve the sum ....
We apply : s = ut + 1/2 atxt in the downward direction...
where :::: s is the distance travelled or displacement
u is the initial velocity which Suzy drops the rock ( 0 m/s1)
t is the time taken ( 0.223 sec )
a is the gravitational acceleration ( 9.8 m/s2 )
So if we apply the equation we get ::
s= ut+ 1/2 atxt
So ,,, s = 0 x 0.223 sec + 1/2 x 9.8 m/s2 x 0.223 sec x 0.223 sec
We get :: s = 0.24367 m as the distance travelled.
So when it was freely dropped from the roof, the rock has travelled a distance of 0.24367 m in 0.223 sec due to the gravitational acceleration of 9.8 m/s2.
So, the height of the roof from the window is 0.24367 m.
The height of the roof from the ground is = 8.2 m + 0.24367 m
= 8.44367 m.
Hope this helps !!
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General Physics Help
Suzy drops a rock from the roof of her house. Mary sees the rock pass her 8.2 m tall window
in 0.223 sec. The acceleration of gravity is 9.8 m/s2 . From how high above the top of the window
was the rock dropped? Answer in units of m.
in 0.223 sec. The acceleration of gravity is 9.8 m/s2 . From how high above the top of the window
was the rock dropped? Answer in units of m.
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| September 10, 2009 04:14 AM |
We apply : s = ut + 1/2 atxt in the downward direction...
where :::: s is the distance travelled or displacement
u is the initial velocity which Suzy drops the rock ( 0 m/s1)
t is the time taken ( 0.223 sec )
a is the gravitational acceleration ( 9.8 m/s2 )
So if we apply the equation we get ::
s= ut+ 1/2 atxt
So ,,, s = 0 x 0.223 sec + 1/2 x 9.8 m/s2 x 0.223 sec x 0.223 sec
We get :: s = 0.24367 m as the distance travelled.
So when it was freely dropped from the roof, the rock has travelled a distance of 0.24367 m in 0.223 sec due to the gravitational acceleration of 9.8 m/s2.
So, the height of the roof from the window is 0.24367 m.
The height of the roof from the ground is = 8.2 m + 0.24367 m
= 8.44367 m.
Hope this helps !!
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