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fiasKO [112]
3 years ago
5

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 se

conds 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 ___.
Physics
1 answer:
White raven [17]3 years ago
3 0

Answer:

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

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From the question,

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7.96 = 7.2 + (<em>n</em> × 0.38)

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P(X < <em>μ </em>+ 2<em>σ</em>) = P(X < <em>μ</em>) + P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>)

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<em />

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K is the kinetic energy.

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With this we have \frac{m}{2} *v^{2}  = m*g*h

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Because when we initially drop the ball, all its energy is potential (and P = - m*g*h) and when it hits the water, all its energy is kinetic (K=\frac{m}{2} *v^{2}. And all that potential was converted to kinetic energy.

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