Question

An aeroplane is flying horizontally with a velocity of 600kmph at a height of 1960m. when it is vertically above a piont A on the ground, a bomb is released from it, the bomb strikes the piont B. find the distance AB. ​

Answer 1

★Given:-

• An aeroplane is flying horizontally with a velocity of 600kmph.
• Height = 1960m
• When it is vertically above a point A on the ground, a bomb is released from it.
• The bomb strikes the point B.

★To find:-

• The distance AB.

★Solution:-

When the bomb is dropped,

Initial horizontal velocity = speed of aeroplane

So, the bomb travels forward after falling.

Given,

=>Initial horizontal velocity = 600kmph

→ 600×(5/18)m/s = 166.67m/s

=>Initial vertical velocity = 0

=> Height = 1960m

=>Time taken by bomb to reach the ground = t

We have,

$\displaystyle \leadsto \underline{\boxed{\sf h=\frac{1}{2} gt^2 }}$

From which,

$\leadsto \boxed{\sf t=\sqrt {\frac{2h}{g} }}$

Substituting the values,

$\displaystyle \mapsto \sf t=\sqrt{\frac{2\times 1960}{9.8} } \\\\\mapsto \sf t=\sqrt{\frac{3920}{9.8} }\\\\\mapsto \sf t=\sqrt{400} \\\\\mapsto \underline{\bold{t=20s.}}$

To find the distance AB:

We know,

$\leadsto \underline{\boxed{\sf Distance = speed \times time }}$

Here,

• Speed = 166.67m/s
• Time = 20s
• Distance (AB) = ?

Putting the values,

$\mapsto \sf AB = speed \times time\\\\\mapsto \sf AB = 166.67 \times 20 = 3333.4\\\\\mapsto \boxed{\boxed{\sf AB = 3333.4m}}$

Hence,

The distance AB is 3333.4m.

_______________

Answer 2

Answer:

From h=

2

1

gt

2

we have t

OB

=

g

2h

OA

=

9.8

2×1960

=20s

Horizontal distance AB=vt

OB

=(600×

18

5

m/s)(20s)

=3333.33m=3.33km