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Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 4.20 m/sm/s. Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 52.9 mm above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 4.50 ss after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.

Required:
a. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?
b. Where is Henrietta when she catches the bagels?

Sagot :

Answer:

a)  v₀ₓ = 9.9 m / s, b)  x_woman = 32.7 m

Explanation:

A) In this exercise, the movement of the bagels is parabolic, we find the time it takes to reach the floor.

          y = y₀ + v_{oy} t - ½ g t²

          0 = y₀ + 0 - ½ gt²

          t = [tex]\sqrt{2y_o/g}[/tex]

let's calculate

          t = [tex]\sqrt{2 \ 52.9/9.8}[/tex]

          t = 3,286 s

Now we can analyze how long Henrieta has walked, she has a walking time before the bagel movement begins (t₀ = 4.50 s)

          t_woman = t₀ + t

           t_woman = 4.50 + 3.286

           t_woman = 7.786 s

The distance traveled in this time is

           x_{woman} = v_woman t_woman

            x_{woman} = 4.20 7.786

          x_{woman} = 32.7 m

For her to grab the bagel, the two of them must be at this point

          x_bagel = x_woman

          x_bael = vox t

          v₀ₓ = x_bagel / t

          v₀ₓ = 32.7 / 3,286

           v₀ₓ = 9.9 m / s

b) when catching the bagels this point x_woman = 32.7 m

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