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?

Answer 1

initial speed = 12.96 m/s

Step-by-step explanation:

Given data:

speed 3.95 m/s

height = 39.9 m

we know that time to reached the bottom is

`H =(1)/(2) gt^2`

`t = \sqrt{(2H)/(g)}`

`= \sqrt{(2* 39.9)/(9.8)}`

t = 2.855 sec

total time taken = 2.85 + 6.50 = 9.35 s

total distance at horizontal
`d = 3.95* 9.35 = 36.95 m`

`speed = (d)/(t) = (36.95)/(2.85) = 12.96 m/s`