A person shooting at a target from a distance of 450 metres finds that the sound of the bullet hitting the target comes 1 / 2 seconds after he fired. A person equidistant from the target and shooting point hears the bullet hit 3 seconds after he heard the gun. The speed of sound is ___.

1350 m/s

Explanation:

Speed of bullet

Distance traveled by the bullet = 450 m

Distance traveled by sound = 450 m

Using S= V × t

==> t= S/V

So, time for bullet t1=450/vb

time for sound t2=450/vs

Because it takes 1/2 sec from firing to hearing of sound by the shooter

==>  t1+t2= 1/2 sec

==> 450/vb+450/vs=0.5sec   (1)

Now a person at distance 'x' each from gun and target takes 3 sec to hear sound from firing to hitting the target.

Fire sound duration

Distance =x m

Speed of sound = V

time =T1

==> T1 = x/vs

Bullet hitting + target sound  duration

During this duration bullet travels a distance of 450 m and sound of hitting travels a distance 'x'

time taken by sound = 450/vb

time taken by sound to travel distance 'x'= x/vs

so let T2= 450/vb + x/vs

But all this happens in 3 sec i-e T = 3 sec

and because firing occurs before hitting the target so he hears strike sound in time T = T2-T1= 450/vb + x/vs -x/vs

But T= 3sec

==> 3= 450/vb

==> vb= 1350 m/s

From eq 1 and 2

again using same relation as above, i-e within 3 seconds he watches bullet traveling a distance of 450 m and hearing sound

no. dot product   =0 when vecs at right angles