A body is projected with an initial velocity 20m/s at 60degree to the

1.	A body is projected with an initial velocity 20m/s at 60degree to the horizontal. its velocity after 1 second in the form of vectors is
Answer» Answer: V = 10 i + 7.32 j Given : ⇒ u = 20 m/s ⇒ angle with HORIZONTAL = α = 60° To Find : velocity after 1 second , V = ? Solution : *At the place of projection ,* $u_{x} = 20 cos 60$ = 20 × 1/2 = 10 m/s $u_{y} = 20sin60$ = 20 × $\sqrt{3} /2$ = 10 $\sqrt{3}$ m /s *After 1 second ,* $v_{x} = 10m/s$ [ because x component of velocity remains same ] $v_{y} = u_{y} + a_{y} .t$ $v_{y}$ = 10 $\sqrt{3}$ - 10 $v_{y}$ = 10 ( $\sqrt{3}$ - 1 ) $v_{y}$ = 10 × 0.732 $v_{y}$ = 7.32 m/s so , V in the form of VECTORS = V = 10 i + 7.32 j