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April 19, 2014

April 19, 2014

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

- Physics -
**drwls**, Wednesday, November 21, 2012 at 5:28amWrite 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 2:33pm0

- Physics -
**atal krishna**, Sunday, April 7, 2013 at 2:12amh=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 2:12amh=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 2:12amh=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

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

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 -
**Anonymous**, Wednesday, February 5, 2014 at 9:31pmall wrong answers

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