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). 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.

Answer 1

To find the velocity of the pot as it hits the ground, we can use the equation vb = Lwt + gt^2, where vb is the speed at the bottom of the window, Lw is the vertical length of the window, t is the time the pot is visible, and g is the acceleration due to gravity.

Step-by-step explanation:

To find the velocity of the pot as it hits the ground, we can use the equation vb = Lwt + gt^2, where vb is the speed at the bottom of the window, Lw is the vertical length of the window, t is the time the pot is visible, and g is the acceleration due to gravity. Since the pot is dropped from above, its initial velocity at the top of the window is 0 m/s. We can calculate the time t using the equation hb = 0.5gt^2, where hb is the height of the window above the ground. Rearranging the equation, we have t = sqrt(2hb/g). Substituting this value of t into the equation vb = Lwt + gt^2, we can solve for vb. Finally, to find the velocity vground as it hits the ground, we can use the equation vground = vb + gt.

Answer 2

`\mathbf{v_(ground) = \sqrt{{v^2+2ghb}}}`

Step-by-step explanation:

From the information given:

The avg. velocity post the window is;

`v_(avg) = (L_w)/(t)`

`v_b` = velocity located at the top of the window

`v_b` = velocity situated at the bottom of the window

Using the equation of kinematics:

`v_b = v_t + gt`

Hence,

`v_t = v_b - gt`

To determine the average velocity as follows:

`v_(avg) = (1)/(2) (v_t + v_b)(L_w)/(t)= (1)/(2)(v_b - gt +v_b) \\ \\(L_w)/(t) = v_b - (1)/(2)gt \\ \\ v_b = (L_w)/(t )+ (1)/(2) gt\\ \\ = (1)/(t) \Bigg(L_w + (1)/(2)gt^2 \Bigg) \\ \\`

where;

`v_b` = velocity gained when fallen through the height h.

Similarly, using the equation of kinematics, we have;

`v_b^2 = 2gh \\ \\h = (v_b^2)/(2g)`

`\implies ((L_w + (1)/(2) gt^2_^2)/(2gt^2)`

Thus, the velocity at the ground is;

`v^2_(grround) = v_b^2 + 2ghb`

`\mathbf{v_(ground) = \sqrt{{v^2+2ghb}}}`