Kevin S. Huang

2010: Problem 4

Topic: Kinematics
Concepts: Free fall

Solution:

The time taken for the piano to drop from rest at $h_1$ to the ground is

h_1=\frac{1}{2}gt_{\text{tot}}^2

t_{\text{tot}}=\sqrt{\frac{2h_1}{g}}

The time taken for the piano to drop from rest at $h_1$ to $h_2$ is

h_1-h_2=\frac{1}{2}gt_1^2

t_1=\sqrt{\frac{2(h_1-h_2)}{g}}

Thus, the time taken for the piano (released at $h_1$ ) to drop from $h_2$ to the ground is

t_2=t_{\text{tot}}-t_1=\sqrt{\frac{2h_1}{g}}-\sqrt{\frac{2(h_1-h_2)}{g}}=\sqrt{\frac{2(30\,\mathrm{m})}{10\,\mathrm{m/s^2}}}-\sqrt{\frac{2(30\,\mathrm{m}-14\,\mathrm{m})}{10\,\mathrm{m/s^2}}}=0.66\,\mathrm{s}

so the answer is A.