Question Solved1 Answer What is the solution for this problem? 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). From what height h above the bottom of your window was the flower pot dropped? Express your answer in terms of Lw, t, and g.

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Transcribed Image Text: 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). From what height h above the bottom of your window was the flower pot dropped? Express your answer in terms of Lw, t, and g.
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Transcribed Image Text: 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). From what height h above the bottom of your window was the flower pot dropped? Express your answer in terms of Lw, t, and g.
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Now,When posses top of window, has follen h. hos velocity v/ 2gh)(u hat(A)^(0)=u hat(A)^(2)+2as)thereTime taken T to J^(@) from h to h+Lw, accelerating at g is{:[L_(w)=v(2gh)t+g+ hat(A)^(2)//2quad(s=(1t+at hat(A)^(2)//2):}],[h=(uL omega hat(A)^(2)-4gL ... See the full answer