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?

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Answer 1 · 2015-11-18T01:04:43

$1.05"s"$

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 = $0.47+0.58=1.05"s"$