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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.

Arun
25763 Points
3 years ago
Angle b/w A & B vectors
tan(α) = B sin(θ)A + B cos(θ)
where θ = angle b/w A & B vectors
Case 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 = 20
So n = 5.