Question

Question #b667f

Answer 1

the plane flies at height, `h=3400m`
angle subtended, `theta=30^@`

`tan theta=("distance travelled in 10s")/h`
`tan 30^@ = x/3400`
`x=1/sqrt3 xx 3400 = 0.58 xx 3400 = 1972m`

now, since
`"speed" = ("distance")/("time")`
`= 1972/10 = 197.2m/s`

Answer 2

speed = 196.3 m/s

Explanation:

The basic formula you have to use will be
speed = distance travelled / time taken ;
Here we know the time as 10 second .
We have a distance of 3400 m but that is NOT the distance travelled

It is rather the height of the aircraft directly above the observer/observation point at a time

The aircraft moves from point A to B in 10 seconds creating an angle of seperation 30 degrees at observation point O .

Using basic trigonometry , we can fin distance AB
From triangle ABO ,

`tan 30^o = (AB)/(OA)`
=> `AB = OA * tan30`
=> `AB = 3400 * 0.5773 = 1963`

So distance AB = 1963 , time taken = 10s
Speed of the aircraft = 1963/10 = 196.3 m/s