Homework Help: Physics

Posted by Zalak on Tuesday, November 20, 2012 at 10:39pm.

The Flower Pot: A flower pot falls off a windowsill and falls past the window below. It takes .3 s to travel from the top of the window to the bottom of the window which is 2.2 m tall. From what height above the top of the window did the flower pot fall?

Physics - drwls, Wednesday, November 21, 2012 at 5:28am
Write equations for the times (t1 and t2) that it takes to pass the top and the bottom of the window, in terms of the unknown distance h.

t1 = sqrt(2h/g)
t2 = sqrt[2(h+2.2)/g]

Then, using those equations, write an equation for the time it takes to go from top to bottom of the window (which you know = 0.3s), in terms of the unknown h.

t2 - t1 = 0.3
= sqrt[2(h+2.2)/g] - sqrt(2h/g)

Solve for h.
Physics - Anonymous, Friday, November 23, 2012 at 2:33pm
0
Physics - atal krishna, Sunday, April 7, 2013 at 2:12am
h=ut +1/2at^2.
2.2=u*.3+1/2*10*(.3)^2
therefore,u=2.2-0.45/.3=5.8.
Taking 'u'=0 and v=5.8m/s.using the equation v^2=u^2+2as,we can caculate 's'=v^2/2a=1.682m.
Physics - atal krishna, Sunday, April 7, 2013 at 2:12am
h=ut +1/2at^2.
2.2=u*.3+1/2*10*(.3)^2
therefore,u=2.2-0.45/.3=5.8.
Taking 'u'=0 and v=5.8m/s.using the equation v^2=u^2+2as,we can caculate 's'=v^2/2a=1.682m.
Physics - atal krishna, Sunday, April 7, 2013 at 2:12am
h=ut +1/2at^2.
2.2=u*.3+1/2*10*(.3)^2
therefore,u=2.2-0.45/.3=5.8.
Taking 'u'=0 and v=5.8m/s.using the equation v^2=u^2+2as,we can caculate 's'=v^2/2a=1.682m.
Physics - atal krishna, Sunday, April 7, 2013 at 2:12am
h=ut +1/2at^2.
2.2=u*.3+1/2*10*(.3)^2
therefore,u=2.2-0.45/.3=5.8.
Taking 'u'=0 and v=5.8m/s.using the equation v^2=u^2+2as,we can caculate 's'=v^2/2a=1.682m.
Physics - atal krishna, Sunday, April 7, 2013 at 2:15am
h=ut +1/2at^2.
2.2=u*.3+1/2*10*(.3)^2
therefore,u=2.2-0.45/.3=5.8.
Taking 'u'=0 and v=5.8m/s.using the equation v^2=u^2+2as,we can caculate 's'=v^2/2a=1.682m.

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