Question

A particle is projected from rest from the top of a vertical towe. In the final 2 seconds it moves a distance 3/4th the height of the tower. Find the height of the tower. (g = 10 ms^-1)

Answer 1

Answer of question #77916, Physics - Mechanics - Relativity

A particle is projected from rest from the top of a vertical tower. In the final 2 seconds it moves a distance $3/4$th the height of the tower. Find the height of the tower. $(g = 10 \, \text{ms}^{\wedge} - 2)$

Input Data:

$\frac{dh}{H} = \frac{3}{4}$, where $dh$ is the distance traveled in the last 2 seconds, and $h$ is the height of the tower

$$g = 10 \frac{m}{s^2}$$

Solution:

Let the tower height be:

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

The distance that the body will pass without the last 2 seconds:

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

Then the distance traveled by the body in the last 2 seconds is equal to:

$$dh = H - h;$$ $$\frac{dh}{H} = \frac{H - h}{H} = 1 - \frac{h}{H} = \frac{3}{4};$$ $$\frac{h}{H} = \frac{1}{4}; \quad H = 4h;$$

Substituting the values, we obtain the quadratic equation:

$$\frac{g t^2}{2} = \frac{4g (t - 2)^2}{2};$$ $$3t^2 - 16t + 16 = 0$$

Solving the equation, we get the time of falling from the tower:

$$t = 4s;$$

Whence the height of the tower is:

$$h = \frac{10 * 4^2}{2} = 80m$$

Answer:

Height of the tower is $80m$

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