Kaitlin is going off to her physics class, jogging down the sidewalk at a speed of 3.30m/s . Her husband Scott suddenly realises that she left in such a hurry that she forgot her sandwich, so he runs to the window of their apartment, which is a height 35.0m above the street level and directly above the footpath, to throw it to her. Scott throws it horizontally at a time 5.50s after Kaitlin has passed below the window, and she catches it on the run. You can ignore air resistance.

[1]With what initial speed must Scott throw the sandwich so Kaitlin can catch it just before it hits the ground?

[2]Where is Kaitlin when she catches the sandwich?

To answer the first question, we need to determine the horizontal distance traveled by Kaitlin in the time it takes for Scott to throw the sandwich.

We can start by calculating the horizontal distance using the formula:

horizontal distance = horizontal velocity × time

Kaitlin's horizontal velocity is given as 3.30 m/s, and the time it takes for Scott to throw the sandwich is given as 5.50 s.

Therefore, the horizontal distance traveled by Kaitlin is:

horizontal distance = 3.30 m/s × 5.50 s = 18.15 m

Since the sandwich must be thrown horizontally to reach Kaitlin, its horizontal velocity is the same as Kaitlin's, which is 3.30 m/s.

Now, to answer the first question, we need to find the initial speed at which Scott must throw the sandwich to ensure Kaitlin catches it just before it hits the ground.

We can use the equation for horizontal motion to find this initial speed:

horizontal distance = initial horizontal velocity × time + (1/2) × acceleration × time²

In this case, acceleration in the horizontal direction is zero since there is no horizontal force acting on the sandwich. We can plug in the values we know:

18.15 m = initial horizontal velocity × 5.50 s

Solving for the initial horizontal velocity:

initial horizontal velocity = 18.15 m / 5.50 s ≈ 3.30 m/s

Therefore, Scott needs to throw the sandwich with the same initial horizontal velocity as Kaitlin's, which is 3.30 m/s, to ensure Kaitlin catches it just before it hits the ground.

To answer the second question, we need to determine where Kaitlin is when she catches the sandwich.

Since the sandwich is thrown horizontally at the same height as Kaitlin's position, and she catches it while still jogging, she will be at the same height as the sandwich. Therefore, Kaitlin catches the sandwich at a height of 35.0 m above the street level, directly below the window of their apartment.