A bullet is fired into the air with an initial upward velocity of 80 ft. per second from the top of building 96 feet high.The equation that gives the height of the bullet at time t is h=96 + 80t - 16t^2.At what time the bullet be 192 ft in the air?
2007-08-19 20:30:40 · 4 answers · asked by kamote 1 in Science & Mathematics ➔ Mathematics
1. Replace 192 in for h in your height equation:
192 = 96 + 80t - 16t^2
2. Write in standard form:
16t^2 - 80t + 96 = 0
3. Divide through by 16:
t^2 - 5t + 6 = 0
4. Factor:
(t-2)(t-3) = 0
5. Write down the solutions:
t = 2, 3.
There are two times the bullet is 192 feet in the air: t=2 and t= 3. The first time, the bullet is still going up. The second time, it is coming back down.
2007-08-19 20:43:01 · answer #1 · answered by Anonymous · 0⤊ 0⤋
192 = 96 + 80t - 16t^2 96 = 80t - 16t^2 80t - 16t^2 = 6
t = 1.2
Maybe that's not right. I guess it was 96=(80t -16t)^2
192=96 + 80t , - 16t no I give up, what is it?
I hope there's no math on the test.
Could you do that factoring part a little slower?
2007-08-20 03:54:14 · answer #2 · answered by hb12 7 · 0⤊ 1⤋
96+80t-16t^2=192-96
96+80t-16t^2=96
80t-16t^2=0
16t(5-t)=0
t=0ort=5
so ans is 5seconds
2007-08-20 03:48:26 · answer #3 · answered by niki einstien 2 · 0⤊ 1⤋
pretty damn quick
2007-08-20 03:43:53 · answer #4 · answered by Anonymous · 0⤊ 0⤋