Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 3.95 m/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 39.9 m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 6.50 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.
Part A. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?
Express your answer using three significant figures.
Part B. Where is Henrietta when she catches the bagels?
Express your answer using three significant figures.

If Bruce throws the bagel horizontally, the vertical component of the motion is purely under gravity with initial speed of 0 m/s. Using one of the equations of motion for the vertical motion,

[tex]h=ut+0.5at^2[/tex]

h = 39.9 m (height of the building)

a = 9.8 (acceleration of gravity)

[tex]t=\sqrt{\dfrac{2\times39.9}{9.8}}= 2.85[/tex]

During this time, Henrietta will have travelled another distance e

[tex]e =3.95\times2.85=11.2575[/tex]Therefore, Henrietta would have jogged a total distance of 25.675 + 11.2575 m = 36.9325 m.

The horizontal speed with which Bruce must throw the bagel is the total distance travelled by Henrietta divided by the time it takes to land vertically i.e.

[tex]v=\dfrac{36.9325}{2.85}=12.96[/tex]

Therefore, the speed of throw is 13.0 m/s.

Henrietta was 36.9 m from the window when she caught the bagel.

alokdubeyvidyaatech alokdubeyvidyaatech · Answer 2 · 2025-03-26T00:23:39-04:00

Part A. The speed at which the bagel was thrown horizontally is 12.9 m/s.

Part B. The total distance traveled by Henrietta when she caught the bagel from the window is 36.9325 m.

How do you calculate the speed of the bagel?

Given that, the speed of Henrietta is 3.95 m/s. When Bruce throws the bagel, the time difference between both of them is 6.5 seconds. Hence the distance can be calculated as given below.

Distance D = Speed [tex]\times [/tex] Time

[tex]D = 3.95 \times 6.50[/tex]

[tex]D = 25.675 \;\rm m[/tex]

Part A

Given that Bruce throws the bagel horizontally, then during the motion, the initial speed will be zero and the vertical component will have gravitational acceleration. In this case, the time required in motion can be calculated by the equation given below.

[tex]s = ut + \dfrac {1}{2} at^2[/tex]

Where, s is the distance, a is the acceleration, t is time and u is the initial speed.

For the horizontal motion, s = 39.9 m, g = 9.8 m/s^2 and u = 0 m/s. The time will be,

[tex]39.9 = 0 + \dfrac {1}{2}\times 9.8\times t^2[/tex]

[tex]t = 2.85 \;\rm s[/tex]

The time required to reach the bagels is 2.85 seconds. But in this time, the distance traveled by Henrietta is given below.

[tex]D' = 2.85 \times 3.95[/tex]

[tex]D' = 11.2527 \;\rm m[/tex]

Hence the total distance traveled by Henrietta is given below.

Distance = 25.675 + 11.2527

Distance = 36.9325 m.

The speed of the bagel is calculated as given below.

Speed = Total Distance / Time

[tex]v = \dfrac {36.9325 }{2.85}[/tex]

[tex]v = 12.9 \;\rm m/s[/tex]

Hence we can conclude that the speed at which the bagel was thrown horizontally is 12.9 m/s.

Part B

The total distance traveled by Henrietta is 36.9325 m. At this distance, Henrietta catches the bagel from the window.

To know more about the speed, follow the link given below.

https://brainly.com/question/7359669.