VIM=? , VIO =? , VIG =?
Given: V0=-5√3î-5ȷ̑= VO ﬩ + V0‖
5√3=10 cos 30
Vm=3î
(VIM)﬩ =(VI-VM)﬩ = - (VOM)﬩ =(VM-VO)﬩
(VIM) ﬩= - (VOM)
(VIM) ﬩= (3î+5√3î)
(VIM) ‖= (VOM)‖ =(V0-VM)‖
(VIM) ‖= -5ȷ̑
(VIM) =(VIM) ‖ + (VIM) ﬩
(VIM) =-5ȷ̑+3î+5√3î =(3+5√3)î=5ȷ̑
VIO =VI – VO for this we need first velocity of image
VIM =VI – VM
VIM + VM =VI
(5√3+3)î-5ȷ̑+3î=VI
⇒ VI = (5√3+6)î-5ȷ̑
VIO =VI –VO
VIO = (5√3+6)î-5ȷ̑-(-5√3î-ȷ̑) = 10√3î +6î+0ȷ̑
VIO =(10√3+6)î
VIG =VI - VG
If ground is supposed to be at rest: VG =0
VIG =VI = (5√3+6)î-5ȷ
