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Rain appears to be falling at an angle 37°with vertical to the driver of a car moving with a velocity of 7m/s. When he increases the velocity of car to 25 m/s, the rain again appears to fall at an angle 37° with vertical. If the actual velocity of rain related to ground is 4nm/s then find n.
Angle b/w A & B vectors tan(α) = B sin(θ)A + B cos(θ)where θ = angle b/w A & B vectorsCase 1: Suppose angle b/w A & B is θ, then tan(900+370) = vr sin(θ)7 + vr cos(θ) −cot(370) = vr sin(θ)7 + vr cos(θ) cot(370) = −vr sin(θ)7 + vr cos(θ) ......eq(1)Case 2: tan(90 −370) = vr sin(θ)25 + vr cos(θ) cot(370) = vr sin(θ)25 + vr cos(θ) ...eq(2)from eq...(1) & (2) −vr sin(θ)7 + vr cos(θ) = vr sin(θ)25 + vr cos(θ) vr cos(θ) = −16 ......eq(3)putting the value of ..eq(3) in eq...(1) we get vr sin(θ) = 12 .....eq(4) [cot(370) = 43]from eq.. eq..(3)2 + eq..(4)2 vr 2 = 400 [sin(θ2) + cos(θ2) = 1] vr = 20 m/s .....eq(5)putting the value of ..eq(5) into eq..(3) we get cos(θ) = −45θ = 900+ 370 = 1270 ...eq(6) So the magnitude of velocity of rain vr= 20 m/s and direction with verticle is at 370 angle Hence velocity = 4n = 20So n = 5.
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