Speed of Sound in Air Two seconds after firing a rifle at a target, the shooters hears the impact of the bullet. Sound travels at 1100 feet per second and the bullet at 1865 feet per second. Determine the distance to the target. ( to the nearest foot) Help me solve this. I don't know how to solve this.
the shooter heard the impact after two seconds (t=2.0 sec.)... the sound travels at 1100 feet per second... the distance then is \[distance=rate \times time\]just plug-in the values...
speed of the bullet has no relation to the speed of sound...
is 1384 the ans?
@Orion1213 is 1384 the right ans?
i got 2200... 1100 x 2...
how did you get 1384 @bless ?
t1 = time of the flight of the bullet t2 = time for the sound to travel to the shooter d = distance to target ---------------- t1 = d/1865 t2 = d/1100 ----------- t1 + t2 = 2 d/1865 + d/1100 = 2 1100d + 1865d = 2*1865*1100 2965d = 4103000 d = 1384 feet
@Orion1213 is this correct?
ok i'm sorry the time given is the total time incurred upon firing then hearing... yes you got it right....
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