# Question #7ef68

May 18, 2016

$h = 25 \cdot 5 = 125 \text{ "meters}$

#### Explanation:

$\text{ let x be displacement in the first second}$
$\text{for t=2} \rightarrow 3 x$
$\text{for t=3} \rightarrow 5 x$

$\text{total=} x + 3 x + 5 x = 9 x$

$\text{thus:}$

$x , 3 x , 5 x , 7 x , 9 x$

$\text{total distance=} x + 3 x + 5 x + 7 x + 9 x = 25 x$

$x = \frac{1}{2} \cdot g \cdot {t}^{2} = \frac{1}{2} \cdot 10 \cdot 1 = 5 \text{ displacement for the first second }$

$h = 25 \cdot 5 = 125 \text{ "meters}$

May 18, 2016

$h = 125 , {s}_{3} = 45 , {t}_{f} = 5$

#### Explanation:

The space covered during the fall is given by
$s = h - \frac{1}{2} g \cdot {t}^{2}$
after $3$[s] we have
$e {q}_{1} \to h - {s}_{3} = h - \frac{1}{2} g \cdot {3}^{2}$
Being ${t}_{f}$ the impact time we have
$e {q}_{2} \to 0 = h - \frac{1}{2} g \cdot {t}_{f}^{2}$
and at ${t}_{f} - 1$ we have
$e {q}_{3} \to {s}_{3} = h - \frac{1}{2} g \cdot {\left({t}_{f} - 1\right)}^{2}$
Solving $\left\{e {q}_{1} , e {q}_{2} , e {q}_{3}\right\}$ for $h , {s}_{3} , {t}_{f}$ we obtain
$h = 125 , {s}_{3} = 45 , {t}_{f} = 5$