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
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.
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To find the apparent speed of the rain to the man walking, we can use vector addition.
First, let's break down the velocity of the rain into its horizontal and vertical components.
The vertical component of the rain's velocity is given by:
Vertical component = 5 m/s * sin(30°)
The horizontal component of the rain's velocity is given by:
Horizontal component = 5 m/s * cos(30°)
Now, let's consider the man's velocity. Since the man is walking towards the east, his velocity is entirely in the horizontal direction.
Therefore, the apparent speed of the rain to the man is the vector sum of the vertical component of the rain's velocity and the man's velocity.
Apparent speed of rain to the man = sqrt(Vertical component^2 + Horizontal component^2 + Man's velocity^2)
Plugging in the given values:
Vertical component = 5 m/s * sin(30°) = 2.5 m/s
Horizontal component = 5 m/s * cos(30°) = 4.33 m/s
Man's velocity = 3 m/s
Apparent speed of rain to the man = sqrt((2.5 m/s)^2 + (4.33 m/s)^2 + (3 m/s)^2)
Calculating this value gives:
Apparent speed of rain to the man ≈ 6.45 m/s
To find the apparent speed of rain from the perspective of the man walking, we need to consider the vector addition of the actual velocity of rain and the velocity of the man.
Given:
Actual velocity of rain (v_rain) = 5 m/s
Speed of the man walking (v_man) = 3 m/s
To find the apparent speed, we need to find the resultant velocity (v_resultant) using vector addition.
Step 1: Resolve the velocity of rain into its horizontal and vertical components.
The horizontal component of rain's velocity (v_rx) can be found using trigonometry:
v_rx = v_rain * cos(30°)
= 5 m/s * cos(30°)
≈ 4.33 m/s (rounded to 2 decimal places)
The vertical component of rain's velocity (v_ry) can be found using trigonometry:
v_ry = v_rain * sin(30°)
= 5 m/s * sin(30°)
≈ 2.5 m/s (rounded to 1 decimal place)
Step 2: Add the horizontal component of rain's velocity to the man's velocity (since they are both in the same direction).
v_resultant = v_rx + v_man
= 4.33 m/s + 3 m/s
= 7.33 m/s (rounded to 2 decimal places)
Therefore, from the perspective of the man walking, the rain appears to be falling at a speed of approximately 7.33 m/s towards east.
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.