Question

Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.70 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 36.5 m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 4.00 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance. Wit what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? Where is Henrietta when she catches the bagels?

Answer 1

The velocity is
`v = 6.66 \ m/s`

Henrietta is at distance
`s= 18.17 \ m` from the under the window

Step-by-step explanation:

From the question we are told that

The speed of Henrietta is
`v= 2.70 \ m/s`

The height of the window from the ground is
`h = 36.5 \ m`

Generally the time taken for the lunch to reach the ground assuming it fell directly under the window is

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

=>
`t = \sqrt{(2 * 36.5 )/(9,8) }`

=>
`t = 2.73 \ s`

Generally the time taken for the lunch to reach Henrietta is mathematically represented as

`T = t + t_1`

Here
`t_1` is the time duration that elapsed after Henrietta has passed below the window the value is given as 4 s

Now

`T = 2.73 + 4`

=>
`T = 6.73 \ s`

Generally the distance covered by Henrietta before catching her lunch is

`s= v * T`

=>
`s= 2.70 * 6.73`

=>
`s= 18.17 \ m`

Generally the speed with which Bruce threw her lunch is mathematically represented as

`v = (18.17)/(2.73)`

`v = 6.66 \ m/s`