Question

A man fires a gun standing in front of a cliff and a woman standing in front of the man at a distance of 640 m hears two sounds at an interval of 2 and 4 seconds respectively. Find the distance between the man and the cliff.

Answer 1

Answer:

Let the distance of the cliff from the initial position of the man be d m.

So, the distance traveled by sound in 3 sec=2d m

So, speed of sound, S=

Time

Distance

S=

3

2d

.....(i)

On moving closer to the cliff by a distance of 82.5 m, the distance=2(d−82.5) m

So,

S=

t

d

S=

2.5

2(d−82.5)

=

2.5

2d−2×82.5

.....(ii)

Therefore form equation (i) and (ii)

3

2d

=

2.5

2d−2×82.5

5d=6d−495

d=495 m

Thus it is the distance of the cliff from the initial position of the man.

Now,

S= 3/2×495 =330 m/s

Answer 2

Answer:

Explanation:

Let the distance of the cliff from the initial position of the man be d m.

So, the distance traveled by sound in 3 sec=2d m

So, speed of sound, S=

Time

Distance

S=

3

2d

.....(i)

On moving closer to the cliff by a distance of 82.5 m, the distance=2(d−82.5) m

So,

S=

t

d

S=

2.5

2(d−82.5)

=

2.5

2d−2×82.5

.....(ii)

Therefore form equation (i) and (ii)

3

2d

=

2.5

2d−2×82.5

5d=6d−495

d=495 m

Thus it is the distance of the cliff from the initial position of the man.

Now,

S= 3/2×495 =330

Hope it is helpful