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A flowerpot falls off a windowsill and passes the win- dow of the story below. Ignore air resistance. It takes the pot 0.380 s to pass from the top to the bottom of this window, which is 1.90 m high. How far is the top of the window below the window- sill from which the flowerpot fell?

Respuesta :

Answer:

d = 0.50 m

Explanation:

Let say the speed at the top and bottom of the window is

[tex]v_1 \: and \: v_2[/tex] respectively

now we have

[tex]d = \frac{v_1 + v_2}{2}t[/tex]

[tex]1.90 = \frac{v_1 + v_2}{2} (0.380)[/tex]

[tex]v_1 + v_2 = 10 [/tex]

also we know that

[tex]v_2 - v_1 = 9.8(0.380)[/tex]

[tex]v_2 - v_1 = 3.72[/tex]

now we have from above equations

[tex]v_2 = 6.86 m/s[/tex]

[tex]v_1 = 3.14 m/s[/tex]

now the distance from which it fall down is given as

[tex]v_f^2 - v_i^2 = 2ad[/tex]

[tex]3.14^2 - 0^2 = 2(9.8)d[/tex]

[tex]d = 0.50 m[/tex]

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