Kinematics!

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Ignoring air resistance, how long would it take a particle to drop .03 meters due to gravity?

Guest Oct 13, 2015

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Best Answer

$h(t)=-9.8 \dfrac {t^2}{2}+v_0 t + h_0 \\ \mbox{In this case we have }v_0=0 \mbox{ and we'll let }h_0=0.03 \mbox{ and solve for when }h=0$

$0=-4.9 t^2 + 0.03 \\ 4.9t^2 = 0.03 \\ t^2 = \dfrac{0.03}{4.9} \\ t=\sqrt{ \dfrac{0.03}{4.9}} = 0.078s$

Rom Oct 13, 2015

Rom · Answer 1 · 2015-10-13T01:54:30

$h(t)=-9.8 \dfrac {t^2}{2}+v_0 t + h_0 \\ \mbox{In this case we have }v_0=0 \mbox{ and we'll let }h_0=0.03 \mbox{ and solve for when }h=0$

$0=-4.9 t^2 + 0.03 \\ 4.9t^2 = 0.03 \\ t^2 = \dfrac{0.03}{4.9} \\ t=\sqrt{ \dfrac{0.03}{4.9}} = 0.078s$

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Guest · Answer 2 · 2015-10-13T01:56:21

Use this famous formula: D=1/2.g.t^2, where D=distance the object has fallen, g=9.81m/sec(the pull of gravity), t=time in seconds.