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

Answer:

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

Explanation:

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

so here we have

$L_w = (\frac{v_b + v_o}{2})t$

here we know that

$v_b$ = speed at the bottom of the window

$v_o$ = speed at the top of the window

$v_b + v_o = \frac{2L_w}{t}$

now we will have

$v_b - v_o = gt$

so we can say

$2v_b = \frac{2L_w}{t} + gt$

$v_b = \frac{L_w}{t} + \frac{1}{2}gt$

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

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

$v_f^2 - (\frac{L_w}{t} + \frac{1}{2}gt)^2 = 2(g)(h_b)$

$v_f^2 = (\frac{L_w}{t} + \frac{1}{2}gt)^2 + 2gh_b$

so we have

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