# Thread: At what thime will the bullet be 192 feet in the air?

1. ## At what thime will the bullet be 192 feet in the air?

A bullet is fired into the air with an initial upward velocity of 80 feet per second from the top of a building 96 feet high. The equation that gives the height h of the bullet at any time t is $h=96+80t-16t^2$. At what thime will the bullet be 192 feet in the air?

2. Originally Posted by Alienis Back
A bullet is fired into the air with an initial upward velocity of 80 feet per second from the top of a building 96 feet high. The equation that gives the height h of the bullet at any time t is $h=96+80t-16t^2$. At what thime will the bullet be 192 feet in the air?
It starts at 96ft. It is moving at 80 ftps.

h = 96 + 80t -16t^2

So we need another 96ft to reach. So we need 192 = 96 + 80t - 16t^2

In other words, 96 = 80t - 16t^2

in other words, 96 + 16t^2 = 80t

in other words, 16t^2 = 80t - 96

so.... basically, t^2 = 5t - 6

so essentially, 3^2 = 9 = 5 x 3 - 6

so basically t = 3

so fundamentally .... 196 = 96 + 240 - 144

so ummm...the time is 3 seconds................lol i think

3. Originally Posted by Alienis Back
A bullet is fired into the air with an initial upward velocity of 80 feet per second from the top of a building 96 feet high. The equation that gives the height h of the bullet at any time t is $h=96+80t-16t^2$. At what time will the bullet be 192 feet in the air?
That will happen twice. Once on the way up, and once on the way down.

$-16t^2+80t-96=192$

Simplifying,

$-t^2+5t-6=0$

$t^2-5t+6=0$

Solve the quadratic for for the two times (t).

4. Lol, silly me not seeing that it was a quadratic equation.