Aflower pot falls from a balcony of an apartment building in salem, oregon and injures a person. your job is to determine the apartment from which it originated. video footage from the second floor indicates that it traveled past a 4.000 meter high window in 0.2600 seconds. all floors in the building are 4.000 meters high. from which floor was the flower pot dropped

The pot was dropped from the fourth floor.

Explanation:

The height and velocity of the pot is given by the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity.

v = velocity at time "t".

Let´s place the origin of the frame of reference on the ground. We know that from the second floor (y = 8.000 m) the pot traveled 4.000 m in 0.2600 s. Then:

y = y0 + v0 · t + 1/2 · g · t²

4.000 m = 8.000 m + v0 · 0.2600 s - 1/2 · 9.8 m/s² · (0.2600 s)²

(4.000 m + 1/2 · 9.8 m/s² · (0.2600 s)² -  8.000 m) / 0.2600 s = v0

v0 = -14.11 m/s

This velocity is the velocity of the pot when it is at a height of 8.000 m (on the second floor).

Now let´s find how much time it took the pot to reach that velocity starting from rest (the pot was dropped, not thrown. Then v0 = 0).

v = v0 + g · t

-14.11 m/s = 0 - 9.8 m/s² · t

t = -14.11 m/s / - 9.8 m/s²

t = 1.440 s.

This is the time it takes the pot to reach the second floor (y = 8.000 m). So, let´s calculate y0 using the equation of height:

y = y0 + v0 · t + 1/2 · g · t²     (v0 = 0)

8.000 m = y0 - 1/2 · 9.8 m/s² · (1.440 s)²

8.000 + 1/2 · 9.8 m/s² · (1.440 s)² = y0

y0 = 18.16 m

The pot was dropped from (18.16 m / 4 = 4.54 ) the fourth floor.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.