A Submarine is traveling at 100 meters depth below water at velocity of 6m / sec. It locates enemy warship ahead on surface of water , It can fires a torpedo at speed of 502 m/sec at 45 0 degree angle to horizontal. Mass of torpedo is 2kg & volume 400cc. Find out horizontal distance H of worship from submarine when torpedo should be fired such that it comes out of water & hits the worship from top (ref figure below ).Neglect water resistance, g =10m/sec2 (Ans = 476 meters)

L3 Q 14

T=time, b = buoyancy, a = accn, t = torpedoF = Force ,w = WaterTorpedowarshipH = ?45 06m/secFb=4NVaUpward force in water = Mg - Buoyancy force FbFw = - 210 + 0.4 10 = -16 N , F= mass Acc. aw 2 = -16N . aw = 8.0m/sec22kg400ccaw = Fw M

T=time, b=buoyancy, A=accn, t=torpedo . aw = 8.0m/sec2 Va2 = 2500 - 2 8100 = 900, Va = 30m/sec warshipH = ?45 06m/secve Va2 = U2 - 2 aw s

t=time, b=buoyancy, A=accn, t=torpedo . Va = 30m/secTorpedowarshipH = ?45 06m/secveVaUawVa = U +aw tw , 30 = 50+(-8 tw ), tw =2.5 secV At top most point in air = 0, 0 = Va + ta gFor ta, 0 = 30 +(-10) ta, ta = 3 secNet air time = 2 ta= 2 3=6sec Net travel time of torpado = 8.5 sec

TorpedowarshipH = ?45 0VaHorizontal V of torpedo = 50 + 6 = 56m/secHorizontal distance by torpedo = 56 8.5 = 476 mThis is constant during motion

L3 -- Q15 :A group of military soldiers ready for marching, make queue 1 km long They start marching ahead at constant speed. At the same time a messenger starts from last solder, reaches first solider, gives message & immediately walks back with return message .By the time he reaches last solider the group has marched 1Km.Speed of messenger is different but constant. Find distance traveled by messenger

A group of military soldiers ready for marching, make queue 1 km long They start marching ahead at constant speed. At the same time a messenger starts from last solder, reaches first solider, gives message & immediately walks back with return message .By the time he reaches last solider the group has marched 1Km.Speed of messenger is different but constant. Find distance traveled by messenger

MP1kmNCY kmPY kmMNq1YC meets AC meets BFrom Eq 1 & 2 ,2 Y 2 1Distance by C = 1+2Y = 1+ 2 1/2 = 1+ 21 + Y km

PY kmStep 1: C meets A at point P after A travels y KMTime taken by both C & A to reach point P is same1 + Y km1kmNMBA = first soldier,B = last soldier, C = messenger,,StartingPositionVa m/secVc m/secC1kmA travels Y km C travels 1+y km

P1kmNStep 2: C returns & meets B at point N. By that time Ar B travels 1Km from start pointCqY km1 - Y1-YTime taken by both C & B to reach point N is sameB travels 1-y C travels y

From Eq 1 & 2 , Distance by C = 1+2Y = 1 + 2 1/2 = 1 + 2 Y

MP1kmNCY kmPY kmMNq1YC meets AC meets BFrom Eq 1 & 2 ,2 Y 2 1Distance by C = 1+2Y = 1+ 2 1/2 = 1+ 21 + Y km

PY kmStep 1: C meets A at point P after A travels y KMTime taken by both C & A to reach point P is same1 + Y km1kmNMBA = first soldier,B = last soldier, C = messenger,,StartingPositionVa m/secVc m/secC1kmA travels Y km C travels 1+y km

P1kmNStep 2: C returns & meets B at point N. By that time Ar B travels 1Km from start pointCqY km1 - Y1-YTime taken by both C & B to reach point N is sameB travels 1-y C travels y

From Eq 1 & 2 , Distance by C = 1+2Y = 1 + 2 1/2 = 1 + 2 Y

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