Question

a body falls freely from the top of the tower.If covers 36% of the total height in last second before striking the ground level,the height of the tower is

Answer 1

Question #54140, Physics / Mechanics | Kinematics | Dynamics |

a body falls freely from the top of the tower. If covers 36% of the total height in last second before striking the ground level, the height of the tower is

Solution:

The vertical motion of the object is described by the equation:

$$d = v_0 t + \frac{1}{2} g t^2$$

, where $d$ – the distance traveled during the last second, $v_0$ – the velocity of the object travelled 36% of the total height.

At the same time the 36% of height equals:

$$0.36h = \frac{1}{2} g t_0^2$$

, where $t_0$ – the time of flight.

Thus, $t_0^2 = 0.72h/g$,

and

$$v_0 = g t_0 = g \times (0.72h/g)^{1/2} = (0.72hg)^{1/2}$$

.

Then $d = h - 0.36h = 0.64h = (0.72hg)^{1/2} t + \frac{1}{2} g t^2$.

Since $t = 1$ s, $d = 0.64h = (0.72hg)^{1/2} + \frac{1}{2} g$.

Let’s $h^{1/2} = x$ and $x > 0$. Then, $0.64x^2 - (0.72g)^{1/2}x - \frac{1}{2}g = 0$

$$D = 0.72g + 4 \times 0.64 \times 0.5g = 7.056 + 12.544 = 19.6$$ D^{1/2} = 4.427$,

x = \frac{[(0.72g)^{1/2} \pm 4.427]}{1.28} = \frac{(2.656 \pm 4.427)}{1.28} = 5.53$$

Finally $h = x^2 = 30.62$ m

Answer: The height of the tower is 30.62 m

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