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]
[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]
