An aeroplane is flying horizontally with a velocity of 600 km/h and at a height of 1960 m. When it is vertically at a point A on the ground a bomb is released from it. The bomb strikes the ground at point B. The distance AB is

Hello student,

Convert velocity in horizontal direction, Ux into m/s:
Ux = 600x1000/3600 = 500/3 m/s
Velocity of the bomb during the flight is constant due to inertia. Therefore,
Initial velocity in vertical direction, Uy = 0
Height = 1960 m
Step 2: Find time ‘t’ using the equation [s = ut + 1/2at^2]
Replace distance ‘s’ by height ‘h’ and acceleration ‘a’ by gravity ‘g’
h = 1/2gt^2
t = √(2h)/g = √(2x 1960/9.8) = 20 sec.
Step 3: Find the distance traversed by the bomb in horizontal direction:
Distance AB = vx t = 500/3 x 20 = 10/3 x 10^3m = 3.333 km.

Thanks and Regards

Shaik Aasif

askIItians faculty

Dear student,

Please find the answer to your problem below.

 

Step 1: Convert velocity in horizontal direction, Ux into m/s:

Ux = 600x1000/3600 = 500/3 m/s

Velocity of the bomb during the flight is constant due to inertia. Therefore,

Initial velocity in vertical direction, Uy = 0

Height = 1960 m

Step 2: Find time ‘t’ using the equation [s = ut + 1/2at^2]

Replace distance ‘s’ by height ‘h’ and acceleration ‘a’ by gravity ‘g’

h = 1/2gt^2

t = √(2h)/g = √(2x 1960/9.8) = 20 sec.

Step 3: Find the distance traversed by the bomb in horizontal direction:

Distance AB = vx t = 500/3 x 20 = 10/3 x 10^3m = 3.333 km.

 

Hope it helps.

Thanks and regards,

Kushagra