Question

Question #fdb55

Answer 1

(a)8.34s and (b )699.81m

Explanation:

Horizontal component of the velocity of decoy is
`=303.9xxcos30"km"/"hr"`

Horizontal displacement of decoy from release point is

`=610m=0.61km`

So the time of flight

`T="Horizontal displacement"/"Horizontal component of velocity"`

`=(0.61km)/(303.9xxcos30 "km"/"hr")xx(3600s)/"hr"=8.34s`

Initial vertical component (downward) of the velocity of decoy is
`=303.9xxsin30"km"/"hr"xx(1000m/"km")/(3600s/"hr")=42.21m/s`

Taking acceleration due to gravity `g =10m/s^2`

The net vertical displacement during time of flight which is the

height of the release point is given by

`H=uxxT+1/2xxgxxT^2`
`=42.21xx8.34+1/2xx10xx(8.34)^2=699.81m`