A person drops a penny from a window that is 50 m high. Neglecting air friction, calculate the speed of the penny when it hits the ground

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[tex]\begin{gathered} h=50m \\ g=9.81m/s^2 \\ vf=\text{?} \\ vf=\sqrt{2gh} \\ vf=\sqrt{(2)(9.81m/s^2)(50m)} \\ vf=31.32\text{ m/s} \\ \text{The }speed\text{ is }31.32\text{ m/s} \\ \text{Question 2} \\ h=0.65m \\ vf=0\text{ m/s} \\ a=-9.81\text{ }m/s^2 \\ vo=\text{?} \\ vf^2=vo^2+2ah \\ \text{solving vo} \\ vo^2=vf^2-2ah \\ vo=\sqrt{vf^2-2ah} \\ vo=\sqrt{(0\text{ m/s})^2-2(-9.81\text{ }m/s^2)(0.65m)} \\ vo=\sqrt[]{0m^2\text{/s}^2+12.753\text{ }m^2/s^2} \\ vo=\sqrt[]{12.753\text{ }m^2/s^2} \\ vo=3.57\text{ m/s} \\ \text{The spe}ed\text{ is }3.57\text{ m/s} \end{gathered}[/tex]

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