A helicopter flies south at 185km/hr relative to the ground. The wind is blowing 60km/hr to the west compared to the ground. What is the helicopter's speed to an observer on the ground?
y component = -185
x component = -60
speed = sqrt(185^2 + 60^2)
tan angle south of west = 185/60
angle south of west = 72
72+180 = 252 in math convention counterclockwise from +x axis
270-72 = 198 in navigation convention clockwise from north (about SSW)
To find the helicopter's speed relative to an observer on the ground, we need to use the concept of vector addition. The speed of the helicopter relative to the observer on the ground can be found by considering the horizontal and vertical components separately.
First, let's find the horizontal component:
The helicopter's horizontal speed relative to the ground is 185 km/hr, and the wind is blowing 60 km/hr to the west. Since the wind is in the opposite direction to the helicopter's motion, we subtract the wind speed from the helicopter's speed:
Horizontal component = Helicopter's horizontal speed - Wind speed
= 185 km/hr - 60 km/hr
= 125 km/hr to the east
Next, let's find the vertical component:
The helicopter is only flying south, which means there is no vertical component due to the wind.
Now that we have the horizontal and vertical components, we can use the Pythagorean theorem to find the magnitude of the helicopter's speed relative to the observer on the ground:
Magnitude of the helicopter's speed = √(Horizontal component^2 + Vertical component^2)
= √(125^2 + 0^2)
= √15625
= 125 km/hr
Therefore, the helicopter's speed relative to an observer on the ground is 125 km/hr.