A body of mass 2 kg has an initial velocity of 3 meters per second along OE and it is subjected to a force of 4 N in a direction perpendicular to OE. The distance of the bodyfrom O after 4 seconds will be

Displacement of body in 4 sec along OEsx=vxt=3×4=12mForce along OF (perpendicular to OE)=4 N∴ ay=Fm=42=2m/s2Displacement of body in 4 sec along OF⇒ sy=uyt+12ayt2=12×2×(4)2=16m  As uy=0∴  Net displacement s=sx2+sy2=(12)2+(16)2=20m