Question

A pilot fires a gun during the plane was flying horizontally
at a velocity of 180 km/h and listen a echo of that sound
after 3 seconds. At what height the plane travelled from
ground? (Velocity of sound in air is 340 m/s)​

Answer 1

Answer:

75m

Explanation:

in this question u=0, v= 180km/hr. that means v=50m/s.

t=3s, h=?

first we have to find a. To find a we will use 1st equation of motion. therefore a= v-u/t

=50-0/3 = 50/3m/s^2.

then we have to find height. to find h we will use 2nd equation. of motion. therefore h=1/2 at^2+ut

=(1/2*50/3*3*3)+(0*3)

=75m which is the answer.

Answer 2

Answer:

504.46m (approx)

Explanation:

A is the initial position from where the gun was fired.

B is the final position of the plane where the pilot heard the sound.

Time taken to cover AB = 3seconds

speed of plane = 180km/hr or 50m/s

therefore, AB = 150m

AC is the incident sound wave which reached the pilot's ears. CB is the reflected wave . AC and CB has to be equal since angle of incidence = angle of reflection. Hence this triangle is an isosceles triangle.

CD = height of plane or the normal of the sound waves.

Now, speed of sound = 340m/s

time taken = 3s

therefore total distance (AC+CB) = 3×340

= 1020m

therefore, AC+CB or AC+AC (since they r equal) = 1020m

Therefore, AC = 1020/2 = 510m

therefore, taking triangle ACD,

=> AD = ½AB ( see the properties of the base of an isosceles triangle . Base is equally divided into two parts by the height )

=> AD = 75m (base)

=> AC = 510m (hypotenuse)

let the height (CD) = x

we know, AC² = AD²+CD²

=> 510² = 75²+x²

=> x² = 260100-5625

=> x = √254475

=> x = 504.455m or 504.46m ( approx )

HOPE IT HELPED :)