SOLUTION: A bullet is fired horizontally at a target, and the sound of its impact is hear 1.5 seconds later. If the speed is 3300Ft/sec and the speed of sound i 1100 ft/sec how far away is t

Question 349027: A bullet is fired horizontally at a target, and the sound of its impact is hear 1.5 seconds later. If the speed is 3300Ft/sec and the speed of sound i 1100 ft/sec how far away is the target?
Answer by haileytucki(390) (Show Source): You can put this solution on YOUR website!
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 3300ft/sec and the speed of sound is 1100 ft/sec how far away is the target?
Let the target be at a distance of x feet .
Time taken by bullet to reach the target = x/3300 seconds
Time taken by sound to reach the ear is = x/1100 seconds
The total time taken is 1.5 seconds

(x)/(3300)+(x)/(1100)=1.5
Find the LCD (least common denominator) of (x)/(3300)+(x)/(1100)+1.5.
Least common denominator: 3300
Multiply each term in the equation by 3300 in order to remove all the denominators from the equation.
(x)/(3300)*3300+(x)/(1100)*3300=1.5*3300
Simplify the left-hand side of the equation by canceling the common factors.
4x=1.5*3300
Multiply 1.5 by 3300 to get 4950.
4x=4950
Divide each term in the equation by 4.
(4x)/(4)=(4950)/(4)
Simplify the left-hand side of the equation by canceling the common factors.
x=(4950)/(4)
Simplify the right-hand side of the equation by simplifying each term.
x=(2475)/(2)
So, 2475/2=(2475)/(2)
The approximate value of (2475)/(2) is 1237.5.
1237.50 feet (distance of the target)
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