Question

A hot air balloon is flying at a constant speed of 20 mi/h at a bearing of N 36° E. There is a 10mi/h cross wind blowing due east. What is the balloon's actual speed and direction? Round angles to the nearest degree and other values to the nearest tenth.

Answer 1

Answer:

The velocity, v = 27.1° mi/h

The direction, θ = 37°NE

Step-by-step explanation:

The velocities are acting on the North and East axis.

North-component or x component:

The North-component of the velocity is given as

$v_{E} = v_{x} = 20sin36 + 10sin90\\\\v_{x} = 11.76 + 10\\\\v_{x} = 21.76 mi/h$

East component or y-component:

The East-component of the velocity is:

$v_{N} = v_{y} = 20cos36 + 10cos90\\\\v_{y} = 16.18 + 0\\\\v_{y} = 16.18 mi/h$

We can then find the resultant velocity:

$v = \sqrt{{v_{x}}^{2} + {v_{y}^{2} }$

$v = \sqrt{21.76^{2} + 16.18^{2}}$

$v = \sqrt{735.29}$

$v = 27.12 mi/h$

v = 27.1° to the nearest tenth.

The direction of the velocity is given as:

θ = $tan^{-1} (\frac{v_{N}}{v_{E}} )$

θ = $tan^{-1} (\frac{16.18}{21.76} )$

θ = $tan^{-1} (0.7436})$

θ  = 36.63° NE = 37° to the nearest degree.

The actual speed of the balloon is 30mi/h and it is moving 37° NE.