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You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?



Answer :

Equation of motion:

[tex] y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2} [/tex]

Since initial velocity is zero, the second term goes away:

[tex]y_{f}=y_{o}+0+ \frac{1}{2} at^{2}[/tex]

[tex]y_{f}=y_{o}+\frac{1}{2} at^{2}[/tex]

[tex]y_{f}-y_{o}= \frac{1}{2} at^{2}[/tex]

[tex]y_{f}-y_{o}=5.4m[/tex]

[tex]5.4m= \frac{1}{2} at^{2}[/tex]

[tex]\frac{2(5.4m) }{a} = t^{2}[/tex]

[tex]a = g = 9.81 \frac{m}{ s^{2}} [/tex]

[tex]\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}[/tex]

[tex]1.1 s^{2} = t^{2}[/tex][tex] \sqrt{1.1 s^{2}} = \sqrt{t^{2}} [/tex]

[tex]t = 1.05s[/tex]

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