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An apple with mass 1.2kg falls out of a tree. Just before it hits the ground, it has a velocity of 7.3m/s. From what height did the apple fall?

Sagot :

xKelvin

Answer:

2.72 m.

Explanation:

Apply conservation of mechanical energy:

[tex]\displaystyle U_i + K_i = U_f + K_f[/tex]

There is zero kinetic energy initially and zero potential energy as the apple hits the ground. Thus:


[tex]\displaystyle U_i = K_f[/tex]

Substitute and solve for h:

[tex]\displaystyle \begin{aligned} mgh_i & = \frac{1}{2}m{v_f}^2 \\ \\ h_i & = \frac{{v_f}^2}{2g} \\ \\ & = \frac{(7.3\text{ m/s})^2}{2(9.8\text{ m/s$^2$})} \\ \\ & \approx 2.72 \text{ m}\end{aligned}[/tex]

Therefore, the apple fell from an approximate height of 2.72 m.

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