Question

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?

Answer 1

`1.05"s"`

Explanation:

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

`v=u+at_1`

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

`:.0=15-32.2t_1`

`:.t_1=15/32.2=0.47"s" " "color(red)((1))`

Now the dog has reached his/her peak height.

How high has the dog jumped?

I'll use `v^2=u^2+2as`

`:.0=15^2-2xx32.2xxs`

`s=225/64.4=3.5"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=1/2"g"t_2^2`

The dog falls 2 + 3.5ft = 5.5ft

`:.5.5=1/2xx32.2xxt_2^2`

`t_2^2=(2xx5.5)/(32.2)`

`t_2=0.58"s" " "color(red)((2))`

Now to get the total time for the dog's journey we add `color(red)((1))` to `color(red)((2))rArr`

Total time = `0.47+0.58=1.05"s"`