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A flowerpot falls from a windowsill 25.0 m above the sidewalk. a. How fast is the flowerpot moving when it strikes the ground? b. How much time does a passerby on the sidewalk below have to move out of the way before the flowerpot hits the ground?

Respuesta :

Answer:

Explanation:

Given

Height of window h=25 m

let us take v be the final velocity of Flower pot before hitting the sidewalk

using [tex]v^2-u^2=2as[/tex]

here [tex]u=initial\ speed =0 m/s[/tex]

thus

[tex]v^2-0=2\times 9.8\times 25[/tex]

[tex]v=\sqrt{490}[/tex]

[tex]v=22.13 m/s[/tex]

Time taken by Flower Pot to reach side walk is

[tex]s=ut+\frac{at^2}{2}[/tex]

[tex]25=0+\frac{9.8\times t^2}{2}[/tex]

[tex]t=\sqrt{\frac{50}{9.8}}[/tex]

[tex]t=2.25 s[/tex]

Thus Passerby has 2.25 s to move out of the way to avoid Flower pot

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