As you look out of your dorm window, a flower pot suddenly falls past. The pot i

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 (h_b) above the ground, what is the velocity (v-gound) of the pot as it hits the ground? You may introduce the new variable (v_b) , the speed at the bottom of the window, defined by (v_b)= (L_w) / t + gt/2 Express your answer in terms of some or all of the variables (h_b) , (L_w) , (t) , (v_b), and g. What is v_ground ?

Explanation / Answer

When passes top of window, has fallen h. Has velocity v(2 g h) (v² = u² + 2 a s) Takes time t to go from h to h + L_w, accelerating at g L_w = v(2 g h) t + g t²/2 (s = u t + a t²/2) h = (4L_w² - 4 g L_w t² + g²t^4)/(8 g t²) Simplifying h = (g t² - 2 L_w)²/(8 g t²)

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