# Question #5d867

Jun 21, 2015

The answer is b) $2 \sqrt{2}$.

#### Explanation:

You know that your object covers half of the height of the tower in 2 seconds. Assuming that this distance is actually the first half of the height of the tower, you can write, taking into consideration the fact that the initial velocity of the object is zero,

${h}_{1} = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{\text{=0}}} \cdot {t}_{1} + \frac{1}{2} g \cdot {t}_{1}^{2}$

${h}_{1} = \frac{1}{2} g \cdot {t}_{1}^{2}$, where

${h}_{1}$ - the first half of the distance;
${t}_{1}$ - the time it needed to cover this distance, in your case ${t}_{1} = \text{2 s}$

The total distance the object travelled can be written as

$H = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{\text{=0}}} \cdot {t}_{T} + \frac{1}{2} \cdot g \cdot {t}_{T}^{2}$, where

${t}_{T}$ - the toal time the object took to reach ground level.

You know that

$H = {h}_{\text{first half" + h_"second half}}$

$H = 2 \cdot {h}_{\text{first half}} = 2 \cdot {h}_{1}$

This means that you have

$2 \cdot {h}_{1} = \frac{1}{2} \cdot g \cdot {t}_{T}^{2}$

$\cancel{2} \cdot \frac{1}{\cancel{2}} \cdot \cancel{g} \cdot {t}_{1}^{2} = \frac{1}{2} \cdot \cancel{g} \cdot {t}_{T}^{2}$

This is equivalent to

${t}_{T}^{2} = 2 \cdot {t}_{1}^{2} = 2 \cdot {2}^{2} = 8$

Thus,

${t}_{T} = \sqrt{8} = \textcolor{g r e e n}{2 \sqrt{2}}$