Question

Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 3.05 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 40.8 mm above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 7.00 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?

Answer 1

v = 10.46 m/s

x = 30.134 m from house

Step-by-step explanation:

given data

speed = 3.05 m/s

time = 7 s

height = 40.8 mm

solution

we get here first time required to fall that is t

t =
`\sqrt{(2* 40.8)/(9.8)}` ..................1

t = 2.88 s

now we take here initial speed that is v

so v × t = 3.05 × ( t+ 7)

v =
`(3.05* (2.88 + 7))/(2.88)`

v = 10.46 m/s

and

when she catch the bagel henrietta was at

x = 3.05 × ( 2.88 + 7)

x = 30.134 m from house