A model water rocket is launched with an initial velocity of 180ft/sec. Its height, h, in feet, after t seconds is given by the formula
h = 180t - 16t(squared)
1. After how many seconds will the rocket first reach a height of 464 ft?
2. After how many seconds from launching will the rocket again be at that same height of 464 ft?
2007-06-29 01:15:14 · 2 answers · asked by andy s 2 in Science & Mathematics ➔ Mathematics
I think you should have chosen the first guy as Best Answer already... Its been an hour...
For your question, I think its all stated in the guy's answer...
=)
2007-06-29 02:32:41 · answer #1 · answered by Popo B 3 · 0⤊ 0⤋
h = 180t - 16t²
464 = 180t - 16t²
This is a quadratic equation, let's set it equal to zero and multiply thru by -1.
16t² - 180t + 464 = 0
divide thru by 4 to simplify more
4t² - 45t + 116 = 0
We will use the quadratic formula to solve.
at² + bt + c = 0
t = (-b ± â(b² - 4ac))/2a
t = (45 ± â(2025 - 1856))/8
t = (45 ± â(169))/8
t = (45 ± 13)/8
t = 4 sec, t = 7.25 sec
.
2007-06-29 08:31:15 · answer #2 · answered by Robert L 7 · 1⤊ 0⤋