Question

A body dropped from the top of the tower clears 9/25th of the total height in its last second of flight.the height of the tower is ? (g=9.8m) and the answer is 122.5.

Answer 1

Answer on Question #44333, Physics, Mechanics | Kinematics | Dynamics

A body dropped from the top of the tower clears 9/25th of the total height in its last second of flight. The height of the tower is ? (g=9.8 m/s²) and the answer is 122.5.

Solution:

An object in free fall experiences an acceleration g of 9.8 m/s². (The - sign indicates a downward acceleration.)

The kinetic equation is

$$y = y _ {0} + v _ {o} t + \frac {1}{2} a t ^ {2}$$

where

$y_0 = h = ?$ is initial position

$v_{0} = 0\,m/s$ is initial speed

$a = g = 9.8\,m/s^2$ is acceleration

At time $t$ the position of a ball is $y = 0$,

and at time $t - 1$ s the position of a ball is $y = \left(1 - \frac{9}{25}\right)h = \frac{16}{25} h$

Thus,

$$\frac {9}{25} h = v _ {o 1} t + \frac {1}{2} g t ^ {2},$$

where $t = 1$ second. We need to find $v_{o1}$, the initial speed as the body enters that last 9/25 of h the height of the tower.

And,

$$v _ {o 1} = \sqrt {2 g \frac {1 6}{2 5} h}$$

assuming the drop means no initial speed at the top. Note 16/25h is the height the body dropped up to the last second.

So,

$$\frac {9}{25} h - \sqrt {2 g \frac {1 6}{2 5} h} - \frac {1}{2} g = 0$$ $$\frac {9}{2 5} h - \frac {4}{5} \sqrt {2 g h} - 4. 9 = 0$$

We define $h = x^2$ so we rewrite to quadratic equation:

$$\frac {9}{2 5} x ^ {2} - \frac {4}{5} \sqrt {2 \cdot 9 . 8} x - 4. 9 = 0$$

Thus,

$$0. 3 6 x ^ {2} - 3. 5 4 1 7 5 x - 4. 9 = 0$$

which we solve for $x$.

$$x_{1,2} = \frac{3.54175 \pm \sqrt{3.54175^2 + 4 \cdot 0.36 \cdot 4.9}}{2 \cdot 0.36}$$ $$x_1 = 11.068$$ $$x_2 = -1.22977$$

and $h = x^2 = 11.068^2 = 122.5 \, \text{m}$.

Answer: $h = 122.5 \, \text{m}$.

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