1D60-001
Bomberdroppingbomb
Projectilemotion
PROMPTSAFTER
STATEMENTOFTHEPROBLEM
Avintagebomberisparticipatinginanairshowandplanstoconductabombingrunusing
an“explosive”flourbag.Iftheaircraftisinlevel-flightat75m/sandreleasestheflour
bomb1500mabovetheground,howfarbackfromthetargetmustthepilotreleasethis
flourbomb(thereleasedistancex)?Howfastisthebombmovinginthehorizontaland
verticaldirectionswhenithitstheground?
STRATEGY
Thisisaprojectilemotionproblemsotheonlyaccelerationaffectingthemotionis
assumedtobeduetogravity,actingverticallydownward.Weneedtoapplythebasic
kinematicsequationsseparatelyforthemotioninthehorizontal,x-direction,andthe
vertical,y-direction.Inthisproblemthereisnohorizontalacceleration,sotherelease
distancewillbethehorizontalvelocitytimestheflighttime.Theflighttimewillcomefrom
analyzingtheverticalmotion-knowingthetotaldistanceandthebomb’sinitialvertical
!
velocity.Thekinematicsequationswewillneedare𝑣 = 𝑣! + 𝑎𝑡and𝑥! = 𝑥! + 𝑣! 𝑡 + 𝑎𝑡 ! .
!
Theseequationscanbewrittenformotioninboththex-andy-directionswiththeflight
time,t,beingacommonvariable.
IMPLEMENTATION
First,weneedtoestablishanoriginandcoordinate
system.Let’ssettheoriginatthepointofreleaseofthe
bombwiththex-axispointedtotherightandthey-axis
pointedup(inastandardconfiguration).
Now,wewilldeterminetheflighttime(i.e.thetimethat
theflourbombtravelsfromreleasetoimpact).The
aircraftisflyinginlevelflight,sotheinitialvelocityinthe
y-directionv0yiszero.Also,theonlyaccelerationisdueto
gravity,actinginadownwarddirection(g=-9.8m/s2).
!
Wewillmanipulate𝑦! = 𝑦! + 𝑣!" 𝑡 + 𝑎! 𝑡 ! andsolvefor
!
flighttime.Notethatusingouroriginsetatthepointof
release,thefinalposition(yf)willbeanegative1500m.
Nowthatwehavetheflighttime,wewilluse𝑥! = 𝑥! +
!
y
v0x
xf
(x0, y0)
yf
x0= y0 = 0.0 m
v0y = 0.0 m/s
g = -9.8 m/s2
yf = -1500 m
v0x = 75 m/s
t, xf, vfy, & vfx = ?
𝑣!" 𝑡 + 𝑎! 𝑡 ! tosolveforthereleasedistanceand𝑣 = 𝑣! + 𝑎𝑡todeterminethevertical
!
speedofthebomb.Thereisnohorizontalacceleration,sothehorizontalvelocityofthe
bombisthesameaswhenitwaspartoftheaircraft.
x
1D60-001
Bomberdroppingbomb
Projectilemotion
PROMPTSAFTER
CALCULATION
First,determinetheflighttimeintheverticaldirection.
!
Startingwith𝑦! − 𝑦! = ∆𝑦 = 𝑣!! 𝑡 + 𝑎! 𝑡 ! ,weget
!
1
−9.8 m/s ! 𝑡 ! 2
= 𝑡 ! and 𝑡 = 17.5 s..
−1500 m − 0 m = 0 +
2 −1500 m / −9.8
!
!!
Next,determinethehowfarbackthebombisreleasedinthehorizontal.
!
!
Startingwith𝑥! − 𝑥! = 𝑣!! 𝑡 + 𝑎! 𝑡 ,weget∆𝑥 = 75
!
!
!
17.5 s + 0 = 1312 m.
Finally,solveforthefinalbombvelocityinthevertical:
Startingwith𝑣!" = 𝑣!! + 𝑎! 𝑡 and 𝑣!" = 0 + −9.8 m/s ! 17.5 s = −171.5 m/s
Inaddition,𝑣!" = 75 m/ssincethereisnoaccelerationinthehorizontal.
SELF-EXPLANATIONPROMPT
1. Whyisthefinalposition,yf,anegativequantity?
2. Whatwouldbedifferentifweweretodesignatetheoriginatthegroundlevelbelowthe
releasepoint?
3. Whatwouldhappeniftheinitialvelocity,v0y,intheverticalwasnotzero?