# A dog leaps off of the patio from 2 feet off of me ground with an upward velocity of 15 feet per second. How long will the dog be in the air?

Nov 18, 2015

$1.05 \text{s}$

#### Explanation:

For the 1st leg of the dog's journey we can use:

$v = u + a {t}_{1}$

I'll use 32.2 ft/s/s for g.

$\therefore 0 = 15 - 32.2 {t}_{1}$

$\therefore {t}_{1} = \frac{15}{32.2} = 0.47 \text{s" " } \textcolor{red}{\left(1\right)}$

Now the dog has reached his/her peak height.

How high has the dog jumped?

I'll use ${v}^{2} = {u}^{2} + 2 a s$

$\therefore 0 = {15}^{2} - 2 \times 32.2 \times s$

$s = \frac{225}{64.4} = 3.5 \text{ft}$

Now for the 2nd leg of the dog's journey. I'll assume he/she falls to the ground below the patio.

I'll use:

$s = \frac{1}{2} \text{g} {t}_{2}^{2}$

The dog falls 2 + 3.5ft = 5.5ft

$\therefore 5.5 = \frac{1}{2} \times 32.2 \times {t}_{2}^{2}$

${t}_{2}^{2} = \frac{2 \times 5.5}{32.2}$

${t}_{2} = 0.58 \text{s" " } \textcolor{red}{\left(2\right)}$

Now to get the total time for the dog's journey we add $\textcolor{red}{\left(1\right)}$ to $\textcolor{red}{\left(2\right)} \Rightarrow$

Total time = $0.47 + 0.58 = 1.05 \text{s}$