document.write( "Question 158515: A bullet if fired horizontally at a target, and the sound of its impact is heard 1.5 seconds later. If the speed of the bullet is 3300ft/sec and the speed of sound is 1100ft/sec, how far away is the target? \n" ); document.write( "
Algebra.Com's Answer #116809 by jojo14344(1513)\"\" \"About 
You can put this solution on YOUR website!
Remember: \"Speed=Distance%2Ftime\" ----> \"S=d%2Ft\", working eqn
\n" ); document.write( "\"Speed%5Bb%5D=S%5Bb%5D=3300%28ft%2Fsec%29\"
\n" ); document.write( "\"Speed%5Bs%5D=S%5Bs%5D=1100%28ft%2Fsec%29\"
\n" ); document.write( "Note: \"d\" in question?
\n" ); document.write( "In our working eqn, we get:
\n" ); document.write( "\"d=S%2At\", where \"S=S%5Bb%5D-S%5Bs%5D\",---> why?
\n" ); document.write( "When the target was hit, sound travelled back 1.5 sec later. It does not mean you hit the target in 1.5 sec, you hit earlier than that. We need to subtract the speed of sound for that matter. Continuing:
\n" ); document.write( "\"S=%283300-1100%29%28ft%2Fsec%29=2200%28ft%2Fsec%29\"
\n" ); document.write( "Then,
\n" ); document.write( "\"d=%282200%28ft%2Fcross%28sec%29%29%29%281.5cross%28sec%29%29\"
\n" ); document.write( "\"d=3300ft\" ---------------------> total distance of the target
\n" ); document.write( "Thank you,
\n" ); document.write( "Jojo\r
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