Question

A particle is dropped vertically from rest, from a height. Find the time taken by
it to fall through successive distances of 1m each.

Answer 1

Answer on Question #43152, Physics, Mechanics | Kinematics | Dynamics

A particle is dropped vertically from rest, from a height. Find the time taken by it to fall through successive distances of 1m each.

Solution.

From relations of uniformly accelerated motion:

$$l = V _ {0} t + \frac {a t ^ {2}}{2}$$

We have $V_{0} = 0$ , as the particle initially was at the state of rest. So for the first segment we have:

$$\Delta h = \frac {g \Delta t _ {1} ^ {2}}{2} \Rightarrow \Delta t _ {1} = \sqrt {\frac {2 \Delta h}{g}}$$

For the first and second segments together:

$$\begin{array}{l} 2 \Delta h = \frac {g (\Delta t _ {1} + \Delta t _ {2}) ^ {2}}{2} \Rightarrow \Delta t _ {2} = \sqrt {\frac {2 * 2 \Delta h}{g}} - \Delta t _ {1} = \sqrt {\frac {4 \Delta h}{g}} - \sqrt {\frac {2 \Delta h}{g}} \\ = \sqrt {\frac {2 \Delta h}{g}} \left(\sqrt {2} - 1\right) \\ \end{array}$$

For the first three segments together:

$$\begin{array}{l} 3 \Delta h = \frac {g \left(\Delta t _ {1} + \Delta t _ {2} + \Delta t _ {3}\right) ^ {2}}{2} \Rightarrow \Delta t _ {3} = \sqrt {\frac {2 * 3 \Delta h}{g}} - \Delta t _ {2} - \Delta t _ {1} \\ = \sqrt {\frac {4 \Delta h}{g}} - \sqrt {\frac {2 \Delta h}{g}} (\sqrt {2} - 1) - \sqrt {\frac {2 \Delta h}{g}} = \sqrt {\frac {2 \Delta h}{g}} (\sqrt {3} - \sqrt {2}) \end{array}$$

And so on. As one can see from relations for $\Delta t_1$ , $\Delta t_2$ and $\Delta t_3$ the relation for interval number $i$ will be next:

$$\Delta t _ {i} = \sqrt {\frac {2 \Delta h}{g}} \left(\sqrt {i} - \sqrt {i - 1}\right)$$

Numerically:

$$\Delta t _ {1} = \sqrt {\frac {2 \Delta h}{g}} = \sqrt {\frac {2 \cdot 1 m}{1 0 \frac {m}{s ^ {2}}}} = \sqrt {\frac {1}{5}} s \approx 0. 4 5 s$$ $$\Delta t _ {2} = \sqrt {\frac {2 \Delta h}{g}} (\sqrt {2} - 1) = \sqrt {\frac {2 \cdot 1 m}{1 0 \frac {m}{s ^ {2}}}} (\sqrt {2} - 1) \approx 0. 1 9 s$$ $$\Delta t _ {3} = \sqrt {\frac {2 \Delta h}{g}} (\sqrt {3} - \sqrt {2}) = \sqrt {\frac {2 \cdot 1 m}{1 0 \frac {m}{s ^ {2}}}} (\sqrt {3} - \sqrt {2}) \approx 0. 1 4 s$$ $$\Delta t _ {i} = \sqrt {\frac {2 \Delta h}{g}} \left(\sqrt {i} - \sqrt {i - 1}\right) = \sqrt {\frac {2 \cdot 1 m}{1 0 \frac {m}{s ^ {2}}}} \left(\sqrt {i} - \sqrt {i - 1}\right) \approx 0. 4 5 \left(\sqrt {i} - \sqrt {i - 1}\right) s$$

Answer: $\Delta t_{i} \approx 0.45\left(\sqrt{i} - \sqrt{i - 1}\right)s$

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