Question

An object falls a distance h from rest. If it travels 0.55h in the last 1.00 s, find (a) the time and (b) the height of its fall.

Answer 1

Answer on Question #51372 – Physics, Mechanics – Kinematics – Dynamics:

An object falls a distance $h$ from rest. If it travels $0.55h$ in the last 1.00 s, find (a) the time and (b) the height of its fall.

Solution.

Let $t$ be the time of fall. We have that free fall is a uniformly accelerated motion with an acceleration of gravity. So:

$$h = \frac {g t ^ {2}}{2};$$

An object travels $0.55h$ in the last 1.00 s, so it travels $0.45h$ in the first $t - 1$ s. Hence:

$$0.45h = \frac {g (t - 1) ^ {2}}{2} \Rightarrow h = \frac {g (t - 1) ^ {2}}{0.9};$$

So we have an equation with respect to $t$:

$$\begin{array}{l} \frac {g t ^ {2}}{2} = \frac {g (t - 1) ^ {2}}{0.9} \Rightarrow \frac {t ^ {2}}{2} = \frac {(t - 1) ^ {2}}{0.9} \Rightarrow \frac {9 t ^ {2}}{20} = (t - 1) ^ {2} \Rightarrow \frac {3 t}{\sqrt {20}} = t - 1 \Rightarrow \\ \Rightarrow \frac {3 t}{2 \sqrt {5}} = t - 1 \Rightarrow \left(1 - \frac {3}{2 \sqrt {5}}\right) t = 1 \Rightarrow t = \frac {1}{1 - \frac {3}{2 \sqrt {5}}} \Rightarrow t = \frac {2 \sqrt {5}}{2 \sqrt {5} - 3} \Rightarrow \\ \Rightarrow t = \frac {2 \sqrt {5} (2 \sqrt {5} + 3)}{11} \Rightarrow t = \frac {20 + 6 \sqrt {5}}{11} \approx 3.04 \text{ s}; \end{array}$$

Now find $h$. Assume that $g = 9.8 \, \text{m/s}^2$:

$$h = \frac {g t ^ {2}}{2} = \frac {9.8 \cdot (20 + 6 \sqrt {5}) ^ {2}}{2 \cdot 121} \approx 45.22 \text{ m}.$$

Answer.

$$t = 3.04 \text{ s}, h = 45.22 \text{ m}.$$

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