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).
A. From what height h above the bottom of your window was the flower pot dropped?
B. If the bottom of your window is a height above the ground, what is the velocity of the pot as it hits the ground?

Answer 1

Answer:

a)   v₁ = Lw / t - ½ gt ,      y = √ v₁² / 2g

b) y_total = Lw + √ (v₁² / 2g),    v = √2 g y_total

Explanation:

a) We can solve this exercise with the free fall equations

        y = y₀ + v₁ t + ½ g t²

     

Point y₀ is the upper part of the window and point Lw the lower part, the velocity v₁ is the velocity at the top of the window

        v₁ t = y - y₀ - ½ g t²

        v₁ = (y-y₀) / t - ½ g t

        v₁ = (Lw -0) / t - ½ g t

        v₁ = Lw / t - ½ gt

This is the velocity at the top of the window

now let's work from the top of the window to where it came loose. At the point where the pot is released if speed is zero

           v₁² = v₀² + 2g y

           v₁² = 2 g y

            y = √ v₁² / 2g

b) for ground speed

           v² = v₀² + 2 g y_total

           v = √2 g y_total

the height in this case is the height from where the pot was dropped

           y_total = Lw + √ (v₁² / 2g)

Answer 2

Answer:

stage c

Explanation: