Answer to Question #123739 in Mechanics | Relativity for Brendon

Question

A helicopter flies with an airspeed of 42.5 m/s [W]. If the wind is traveling with a velocity of 25.0 m/s [E30°S] relative to the ground then determine the velocity of the helicopter relative to the ground.

Use the data from the question to determine the velocity components for the helicopter’s speed relative to the air and the wind’s speed relative to the ground.
Use your velocity components from part (a) and determine the resultant velocity for each component. (In other words, determine the resultant horizontal component for the helicopter’s ground speed and the resultant vertical component for the helicopter’s ground speed).
Using your data, determine the resultant velocity of the helicopter relative to the ground.

Answer 1

Explanations & Calculations

• Consider figure (1),
• V _h,a = velocity of helicopter relative to air V_a,e = velocity of air relative to earth
• V _h,e = velocity of helicopter relative to air

• Consider the component wise breakdown of the velocity of the helicopter relative to earth as shown in figure (2).

• Apply the relationship V(h,e) = V(h,a) + V(a,e) along the needed direction to find each of the components

(1).

$\qquad\qquad \begin{aligned} \small \overleftarrow V_{horizontal} &= \small \overleftarrow{42.5}ms^{-1} + \overleftarrow{(-25\sin 30^0)}ms^{-1}\\ &= \small \overleftarrow{42.5}ms^{-1} + \overrightarrow{12.5} ms^{-1}\\ &= \small \overleftarrow{17.5} ms^{-1} \end{aligned}$

(2).

$\qquad\qquad \begin{aligned} \small \downarrow V_{vertical} &= \,\small \downarrow 0ms^{-1} + \downarrow {25\cos 30^0}\\ &= \,\small \downarrow \frac{25\sqrt3}{2}ms^{-1} \end{aligned}$

• Therefore, velocity of the helicopter relative to ground/earth could be calculated as follows.

$\qquad\qquad \begin{aligned} \small V_{h,e} &= \small \sqrt{V_{horizontal}^2 + V_{vertical}^2}\\ &= \small \sqrt{17.5^2 + \big(\frac{25\sqrt3}{2}\big)^2}\\ &= \small \bold{25.84ms^{-1}} \end{aligned}$

$\qquad\qquad \begin{aligned} \small \text{Direction} &= \small \tan^{-1}\Big(\frac{V_{vertical}}{V_{horizontal}}\Big)\\ &= \small \tan^{-1}\Big(\frac{5\sqrt3}{7}\Big)\\ &= \small \bold{51.05^0} \end{aligned}$

• Velocity is, 25.84 ms^-1 [S 51.05⁰ W]