Question

Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 4.15 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 55.2 mm above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 5.50 ss after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.

(a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground?
(b) Where is Henrietta when she catches the bagels?

Answer 1

Step-by-step explanation:

Distance travelled by Henrietta in 5.5 s = 4.15 x 5.5 = 22.825 m .

Time taken by lunch of bagels to fall vertically by 55.2 m . Let it be t .

s = ut + 1/2 g t²

55.2 = 0 + .5 x 9.8 x t²

t² = 11.26

t = 3.356 s

By the time the lunch of bagels touches the hand of Henrietta , she would have travelled further by distance

s = 3.356 x 4.15 = 13.9 m

She is now at distance of 22.825 + 13.9 = 36.725 m from window .

So lunch of bagels must travel a horizontal distance of 36.725 m in 3.356 s which the time of fall of bagel .

Speed of bagel = distance / time

= 36.725 / 3.356

= 10.94 m /s

b )

Henrietta is 36.725 m from window at the time when she catches the bangel.