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?

1 Answer
Nov 18, 2015

Answer:

#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"#