Question 349027
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)