Question 400646
A certain .22 caliber gun fires a bullet at a speed of 1192 feet per second.
 If the .22 caliber gun is fired straight upward into the sky, the height of
 the bullet in feet is given by the equation , h=-16t^2 +1192t where t is the
 time in seconds with t=0 corresponding to the instant the gun is fired.
 How long is the bullet in the air?
:
The bullet will be at h=0 twice, when it is shot and when it returns so we have:
-16t^2 + 1192t = 0
factor out -t, results:
-t(16t - 1192) = 0
t = 0 
and 
16t = 1192
t = {{{1192/16}}}
t = 74.5 seconds