# Distance to target

• Aug 12th 2008, 11:21 AM
sstecken
Distance to target
Question:
Distance to a target: A bullet is fired horizontally at a target, and the sound of its impact is heard 1.5 seconds later. If the speed of the bullet is 3300 ft/sec and the speed of sound is 1100 ft/sec, how far away is the target?

I know that 0.25 of the 1.5 sec journey (0.375sec) is spend on the bullet and that the distance is 1237.5 ft. This corresponds with the back of the book and makes sense. Still, I'm not sure how to formalize this question into a practical equation. Can someone give it a go?

Thanks.
• Aug 12th 2008, 01:06 PM
CaptainBlack
Quote:

Originally Posted by sstecken
Question:
Distance to a target: A bullet is fired horizontally at a target, and the sound of its impact is heard 1.5 seconds later. If the speed of the bullet is 3300 ft/sec and the speed of sound is 1100 ft/sec, how far away is the target?

I know that 0.25 of the 1.5 sec journey (0.375sec) is spend on the bullet and that the distance is 1237.5 ft. This corresponds with the back of the book and makes sense. Still, I'm not sure how to formalize this question into a practical equation. Can someone give it a go?

Thanks.

Let the distance to target be x. Then time of flight of bullet is x/3300 s, and the time for the sound to return is x/1100, so the total time is:

t=x/1100(1/3+1)=x*0.00121212=1.5 s.

so x=1.5/0.00121212=1237.5 ft

RonL
• Aug 12th 2008, 02:40 PM
Soroban
Hello, sstecken!

Quote:

A bullet is fired horizontally at a target, and the sound of its impact is heard 1.5 seconds later.
If the speed of the bullet is 3300 ft/sec and the speed of sound is 1100 ft/sec,
how far away is the target?

$\text{We will use: }\;\text{Distance} \:=\:\text{Speed} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{S}$

The bullet travels $x$ feet at 3300 ft/sec. .This takes: . $\frac{x}{3300}$ seconds.

The sound travels $x$ feet at 1100 ft/sec. .This takes: . $\frac{x}{1100}$ seconds.

. . And the total time is 1.5 seconds.

There is our equation! . $\hdots\quad \frac{x}{3300} + \frac{x}{1100} \:=\:1.5$

Multiply by 3300: . $x + 3x \:=\:4950\quad\Rightarrow\quad 4x \:=\:4950$

Therefore: . $x \;=\;1237.5\text{ ft}$