A stone dropped from the edge of the roof. it passes a window 2 m high in 0.1 sec . how far is the roof above the top of window
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The object is falling under gravity. As it is moving towards the earth, Its acceleration is +g.
As it is dropped, It has initial velocity = 0 .
The stone passes the window of 2m high in 0.1s .
Let the distance between the roof and that window be " k " k
m and the time taken to reach that " k "m is t
That is k = 1/2gt² ( From the second equation of motion)
Given that, Stone passes through the window in 0.1 second.
So it travels k+2 m in t + 0.1seconds .
k + 2 = 1/2g(t+ 0.1)²
Substituting the value of k,
1/2gt² + 2 = 1/2g(t²+0.01 + 0.2t)
1/2gt² - 1/2gt² + 2 = 0.005 g + 0.1gt
2 = 0.005 * 10 + 0.1(10)t
2= 0.05 + t
2 - 0.05 = t
1.95 = t.
Therefore, The stone takes 1.95seconds to travel from roof to top of the window.
Now, We have to find The distance between roof and top of the window ( k)
k = 1/2(10)(t²)
k = 5t²
k = 5(1.95)²
k = 19.0125
Therefore, The height of the roof above the top of the window is 19.0125m
The object is falling under gravity. As it is moving towards the earth, Its acceleration is +g.
As it is dropped, It has initial velocity = 0 .
The stone passes the window of 2m high in 0.1s .
Let the distance between the roof and that window be " k " k
m and the time taken to reach that " k "m is t
That is k = 1/2gt² ( From the second equation of motion)
Given that, Stone passes through the window in 0.1 second.
So it travels k+2 m in t + 0.1seconds .
k + 2 = 1/2g(t+ 0.1)²
Substituting the value of k,
1/2gt² + 2 = 1/2g(t²+0.01 + 0.2t)
1/2gt² - 1/2gt² + 2 = 0.005 g + 0.1gt
2 = 0.005 * 10 + 0.1(10)t
2= 0.05 + t
2 - 0.05 = t
1.95 = t.
Therefore, The stone takes 1.95seconds to travel from roof to top of the window.
Now, We have to find The distance between roof and top of the window ( k)
k = 1/2(10)(t²)
k = 5t²
k = 5(1.95)²
k = 19.0125
Therefore, The height of the roof above the top of the window is 19.0125m
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