Question

A body falls freely from top for a tower . It covers 36% of total height in last second before stricking the ground level. A height of tower is?

Answer 1

Answer on Question #48148 – Physics – Mechanics | Kinematics | Dynamics

1. A body falls freely from top for a tower. It covers 36% of total height in last second before stricking the ground level. A height of tower is?

$$\begin{array}{l} \eta = 0.36 \\ t_0 = 1\,s \\ \hline h - ? \end{array}$$

Solution.

Let introduce the coordinate system, so that $Y$-axis is directed vertically upwards and zero level corresponds to the ground.

The coordinate of a body obeys the following law: $y = h - \frac{gt^2}{2}$,

where $h$ is the tower height.

The total time of falling (at that time $y = 0$): $t_1 = \sqrt{\frac{2h}{g}}$.

According to the task, $y(t_1 - t_0) - y(t_1) = \eta \cdot h$, $\left[h - \frac{g}{2} \left( \sqrt{\frac{2h}{g}} - t_0 \right)^2 \right] - 0 = \eta \cdot h$.

Solving the last equation, one can obtain the height: $h = \frac{g}{2} \left( \frac{t_0}{1 \pm \sqrt{1 - \eta}} \right)^2$.

Let check the dimension: $\left[h\right] = \frac{m}{s^2} \cdot s^2 = m$.

Let evaluate the quantity:

$$h = \frac{9.81}{2} \cdot \left( \frac{1}{1 - \sqrt{1 - 0.36}} \right)^2 = 122.6\,(m), \quad h = \frac{9.81}{2} \cdot \left( \frac{1}{1 + \sqrt{1 - 0.36}} \right)^2 = 1.51\,(m).$$

Answer: $122.6\,m$ or $1.51\,m$.

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