Question

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.

Answer 1

Answer:

Part A 12.9 m/s

Part B 36.9 m

Explanation:

Henrietta ran at 3.95 m/s. She was 6.5 seconds away from from the window when Bruce throws the bagel. This equates to a distance of d

$d= 3.95\times6.5=25.675$ (distance = speed x time)

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,

$h=ut+0.5at^2$

h = 39.9 m (height of the building)

a = 9.8 (acceleration of gravity)

$t=\sqrt{\dfrac{2\times39.9}{9.8}}= 2.85$

During this time, Henrietta will have travelled another distance e

$e =3.95\times2.85=11.2575$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.

$v=\dfrac{36.9325}{2.85}=12.96$

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

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