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 actual speed = 27.12 mi/h

Direction = 36.5° in NE(north of east)

Step-by-step explanation:

As given , A hot air balloon is flying at a constant speed of 20 mi/h at a bearing of N 36° E.

⇒θ = 36°

Let v₀ be the constant speed, then v₀ = 20

Let vₓ be the speed in East direction

     $v_{y}$ be the speed in North direction

So,

vₓ  = v₀ sin(θ) = 20 sin(36°) + 10 ( As given, There is a 10mi/h cross wind blowing due east.)

⇒vₓ  = 20(0.588) + 10 = 11.76 + 10 = 21.76 mi/h

and   $v_{y}$  = v₀ cos(θ) = 20 cos(36°) = 20(0.809) = 16.18 mi/h

Now,

the actual speed = √(vₓ)² + ($v_{y}$)²

                            = √(21.76)² + (16.18)²

                            = √473.498 + 261.792

                            = √735.29 =  27.12

⇒The actual speed = 27.12 mi/h

Now,

Direction = θ = $tan^{-1}(\frac{v_{y} }{v_{x} } ) = tan^{-1}(\frac{16.18 }{21.76 } ) = tan^{-1}(0.74) = 36.5$

⇒ Direction = 36.5° in NE(north of east)