Kinematics
Advanced
linear motion
Source: High School Physics Olympiad
Problem Sets:
kinetics - olympiad
Problem
A freely falling body, released from rest, covers a distance equal to $\frac{1}{3}$ of its total path during the last second of its fall. Take $g = 9.8$ m/s$^2$.
- Find the total time of the fall.
- Find the height from which the body falls.
$t = \dfrac{1}{1 - \sqrt{2/3}} \approx 5.45$ s; $h \approx 145$ m
Let the total fall time be $t$ and the total height $h = \frac{1}{2}gt^2$.
The distance covered in the first $(t - 1)$ seconds is $\frac{1}{2}g(t-1)^2$, which must equal $\frac{2}{3}h$:
$$\frac{1}{2}g(t-1)^2 = \frac{2}{3} \cdot \frac{1}{2}gt^2$$So $(t-1)^2 = \frac{2}{3}t^2$, giving $t - 1 = \sqrt{\frac{2}{3}}\,t$ (the root $t - 1 = -\sqrt{\frac{2}{3}}\,t$ gives $t < 1$ s, which is rejected since the fall lasts longer than its last second).
$$t = \frac{1}{1 - \sqrt{2/3}} \approx 5.45 \text{ s}$$Then the height:
$$h = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(5.45)^2 \approx 145 \text{ m}$$