Question

An aeroplane flying horizontally with a
speed of 366 km/h releases a bomb
at a height of 490m from the
ground. When will the bomb strike the ground?

A)8 sec
B)6 sec
C)7 sec
D)10 sec​

Answer 1

Answer:

From the definition of Time Of Flight

We know that,

T= √2h/g

= √ 2* 490/9.8

=10 sec

Therefore, the Bomb will strike the ground at 10secs

Answer 2

Answer:

# Given:

Speed of the aeroplane horizontally :

= 366 km/hr

Height by which the bomb was released :

= 490 m

Acceleration due to gravity :

= 9.8 m / s

# To find:

How much time the bomb will take to strike the ground

# Taken:

$\bold{t = \sqrt{ \frac{2h}{g} } }$

Where,

t = Time taken

h = Height by which the bomb released

g = Acceleration due to gravity

# Solution:

$\bold \purple{Taken ,\: t = \sqrt{ \frac{2h}{g} } }$

$\bold{t = \sqrt{ \frac{2 \times 490m}{9.8} } }$

$\bold{t = \sqrt{ \frac{980}{9.8} } }$

$\bold{t = \sqrt{ \frac{9800}{98} } }$

$\bold{t = \sqrt{100} }$

$\bold{t = 10 \: seconds}$

# Answer:

So , the time taken by the bomb to reach the ground is 10 seconds.

# Additional information:

If you want to find at what distance will the bomb strike the ground use this formula:

$\bold{V×(\frac{T}{VT})}$

Where,

V = Horizontal velocity of the aeroplane

T = Time taken by the bomb to reach the ground

SOLUTION:

$\bold{366km/hr×(\frac{10}{366×10})}$

$\bold{366km/hr(\frac{10}{3660})}$

$\bold{366km/hr×\frac{1}{366}}$

$\bold{= 1\:km}$

So , the bomb will strike at the distance of 1 km ...

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