03.To a man walking at the rate of 2 km/hour with respect to ground, the rain appears to fall verticallyWhen he increases his speed to 4 km/hour in same direction of his motion, rain appears to meet himat an angle of 45° with horizontal, find the real direction and speed of the rain.
Answers
Answer:
To a man walking at the rate of 2 km/hour with respect to ground, the rain appears to fall verticallyWhen he increases his speed to 4 km/hour in same direction of his motion, rain appears to meet himat an angle of 45° with horizontal
Explanation:
Let i^ and j^ be unit vector in horizontal and vertical respectively.
Let velocity of rain vr=ai^+bj^
speed of rain=√ a^2+b^2−−−−−−
Case I:
when relative velocity of rain w.r.t man is vertical
vrm=v¯r−v¯m
vm=2i^
vrm=(a−2)i^+bj^
Since vrm is vertical
a−2=0=>a=2
Case II:
when relative velocity is at 45∘
v¯m=4i^
vrm=(a−4)i^+bj^
=−2i^+bj^
Since tanθ= b/−2
tan 45=1
=> |b|=2
Therefore Speed=√2^2+2^2
=2√2
Explanation:
Let the velocity of rain be x i + y j,
where i is the direction in which man is walking and j is the vertically upward direction.
Relative velocity of the rain when the man is walking at 2 km/hr is
(x - 2) i + y j
As this is vertically downwards,
x - 2 = 0
x = 2 ……….(1)
Also, the relative velocity of the rain when the man is walking at 4km/hr is
(x - 4) i + y j
=> y/(x - 4) = arctan(45°) = 1
=> y = x - 4 --------------------------------------(2)
Put the value of x from eq (1) into eq (2)
y = x - 4
= 2 - 4
= - 2
=> speed of rain is √[(2)^2 + (-2)^2]
=√4+4
= √8 km/hr
=2√2