Question

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is Lw. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.

Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

Question:

If the bottom of your window is a height hb above the ground, what is the velocity vground of the pot as it hits the ground? You may introduce the new variable vb, the speed at the bottom of the window, defined by

vb=Lwt+gt2.

Express your answer in terms of some or all of the variables hb, Lw, t, vb, and g.

Answer 1

The final velocity of the flower pot as it hits the ground is calculated by taking the square root of the sum of the square of the velocity at the bottom of the window and two times the acceleration due to gravity times the height above the ground.

Step-by-step explanation:

To find the velocity of the flower pot as it hits the ground (vground), we first consider the velocity of the pot as it passes the bottom of the window (vb). The flower pot starts with an initial velocity of zero since it was dropped, not thrown. The velocity at the bottom of the window is given by the equation vb = Lw/t + (g*t)/2, according to the question prompt. We will use this velocity as the initial velocity for the remainder of the pot's fall to the ground.

Next, we apply the equation of motion v2 = vo2 + 2gh to find the final velocity, substituting vo for vb and h for hb, the height above the ground:

vground2 = vb2 + 2g * hb

Solving for vground:

vground = √(vb2 + 2g * hb)

This equation provides the velocity of the flower pot as it hits the ground in terms of the given variables hb, Lw, t, vb, and g.

Answer 2

`v_f = \sqrt{((L_w)/(t) + (1)/(2)gt)^2 + 2gh_b}`

Step-by-step explanation:

As we know that pot is visible to us for "t" time

so here we have

`L_w = ((v_b + v_o)/(2))t`

here we know that

`v_b` = speed at the bottom of the window

`v_o` = speed at the top of the window

`v_b + v_o = (2L_w)/(t)`

now we will have

`v_b - v_o = gt`

so we can say

`2v_b = (2L_w)/(t) + gt`

`v_b = (L_w)/(t) + (1)/(2)gt`

now velocity of the pot just before it hit the floor is given as

`v_f^2 - v_i^2 = 2 a d`

`v_f^2 - ((L_w)/(t) + (1)/(2)gt)^2 = 2(g)(h_b)`

`v_f^2 = ((L_w)/(t) + (1)/(2)gt)^2 + 2gh_b`

so we have

`v_f = \sqrt{((L_w)/(t) + (1)/(2)gt)^2 + 2gh_b}`