<p>A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes 0.434 s to pass this window, which is 2.10 m high. How far is the top of the window below the windowsill from which the flowerpot fell?</p>
<p>You just have to apply the kinematics equations to the top of the window and the bottom of the window and solve a system of equations.</p>
<p>Let the distance between the windowsill and the top of the window below be x, the velocity of the flowerpot at the top of the window be v, height of window be h, time be t,</p>
<p>then v = sqrt( 2gx )</p>
<p>Also, h = vt + (1/2)gt^2
therefore, v= (h - (1/2)gt^2)/t = (2.10 - (.5)(9.8)(.434^2))/.434 = 2.712m/s</p>
<p>thus, sqrt( 2gx ) = 2.712 </p>
<p>Solve for x, we find x=0.375m</p>
<p>ok thanks a lot, i needed to look at my newtonian mechanics part on the equation sheet</p>
<p>whoa that's on my physics summer assignment, wha's ur name again urcool?</p>
<p>Isn't this problem from the GIANCOLI book? I remember posting this exact same problem on here in September, but no one answered me.</p>