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?

Question

contestada

Respuesta :

moya1316 moya1316 · Answer 1 · 2021-03-05T02:24:09+01:00

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