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Answer 1 · 2016-07-16T20:01:47

(a)8.34s and (b )699.81m

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

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