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

27.1 mi / h; N 53° E

Step-by-step explanation

Answer 2

The actual speed = 27.12 mi/h

Direction = 36.5° in NE(north of east)

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)((v_(y) )/(v_(x) ) ) = tan^(-1)((16.18 )/(21.76 ) ) = tan^(-1)(0.74) = 36.5`

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