Asked by shivani
rain is falling at 5m/s making 30 degrees with the vertical towards east . to a man walking at 3m/s on a horizontal road towards east , the rain appears to be at a speed of
Answers
Answered by
Henry
All angles are measured CCW from due East.
Vr = 3 + 5[300o].
Vr = 3 + 5*cos300+5*sin300 = 3 + 2.5 - 4.33i = 5.5 - 4.33i = 7m/s[322o] = Velocity of the rain.
322o CCW = 38o S. of E.
Vr = 3 + 5[300o].
Vr = 3 + 5*cos300+5*sin300 = 3 + 2.5 - 4.33i = 5.5 - 4.33i = 7m/s[322o] = Velocity of the rain.
322o CCW = 38o S. of E.
Answered by
Anand
Rain doesn't appear to fall vertically downwards,so we need to resolve Vr into components and then solve.
Vr/m=Vr - Vm=Vr + (-Vm)
So man's apparent velocity becomes 3m/s towards west.
Vr makes 30° with vertical,so
Vrcos30°=2.5√3 m/s vertically downwards and Vrsin30°=2.5m/s east.
So,net velocity in horizontal direction =
3-2.5=0.5 west.
Finally,Vr/m=√(0.5)^2+(2.5√3)^2.
=√0.25+(6.25x3).
=√0.25+18.75.
=√19.
Please tell me if it was helpful to you😄☺️
Vr/m=Vr - Vm=Vr + (-Vm)
So man's apparent velocity becomes 3m/s towards west.
Vr makes 30° with vertical,so
Vrcos30°=2.5√3 m/s vertically downwards and Vrsin30°=2.5m/s east.
So,net velocity in horizontal direction =
3-2.5=0.5 west.
Finally,Vr/m=√(0.5)^2+(2.5√3)^2.
=√0.25+(6.25x3).
=√0.25+18.75.
=√19.
Please tell me if it was helpful to you😄☺️
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