A projectile is shot vertically in the air from ground level. its height h, in feet, after t seconds is given by h = 88t - 16t^2. Will the projectile have a height of 125 feet 0, 1, or, 2 times after being shot?
also...
find the values for c so that s^2 + cs + 81 =0 has one real root
PLEASE HELP ME!! I'M STRUGGLING WITH THESE TWO MATH PROBLEMS..PLEASE!!
2007-05-24 20:53:38 · 8 answers · asked by eliza 2 in Science & Mathematics ➔ Mathematics
In the first problem, h will be zero twice, once at t=0 and the other time at t=88/16. It will be at it's maximum height at the time in the middle, which happens to be 88/16 divided by 2 or 44/16. Plug in 44/16 into the first formula to find the max height.
88(44/16) - 16(44/16)^2 = 121. So, it will be at 125 feet zero times.
The value of c for which the equation has one real root makes the quadratic a perfect square quadratic or c = -18 and c = 18
In these cases, the quadratic factors as (s-9)^2 and (s+9)^2
P.S. The first answer isn't always the right answer :)
2007-05-24 21:05:57 · answer #1 · answered by Anonymous · 1⤊ 0⤋
dh/dt = 88 - 32t
....... = 0 at highest point
t = 11/4
s = 242 - 121 = 121m
Projectile does reach 125 m so the answer is 0
s² + cs + 81 = 0
"b² - 4ac = 0 " for one real root
so c² - 4 x 81 = 0
c = ± 18
2007-05-24 22:11:11 · answer #2 · answered by fred 5 · 0⤊ 0⤋
Answer 2:
s^2 + cs + 81 =0
For the equation to have one real root, the discriminant must be 0.
i.e. c^2 - 4*81 = 0
c^2 - (2^2 * 9^2) = 0
c^2 - (2 * 9)^2 = 0
c^2 = 18^2
c = 18
When c = 18, the eqn. has one real root.
Another way:
s^2 + cs + 81 =0
Perfect square polynomials, when equated to 0, have one real root.
s^2 + cs + 81 =0
(s)^2 + cs + (9)^2 = 0
It is clear that it is in the form (a + b)^2
where a = s, b = 9
cs = 2ab
cs = 2*9*s
c = 2*9
c = 18
2007-05-24 21:20:24 · answer #3 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 1⤋
1. the equation shows a maximum h of 121 feet. So it will have a h=125 zero times. First, find the vertex is with the vertex formula tmax=-b/(2a) or tmax=-88/(2*-16) or tmax=2.75
Plug this t into the formula h=88*2.75-16*2.75^2.75 hmax=121
2007-05-24 21:20:31 · answer #4 · answered by Anonymous · 0⤊ 1⤋
h' = 88 - 32 t
max height at t = 88/32 = 2.75
h max = (88 - 16 * 2.75)*2.75 = 77 * 2.75 = 211.75
so it must have height 125 twice
(once on way up, once on way down)
s^2 +/- 18s + 81 = (s+/-9)^2
so c = +/-18
2007-05-24 21:06:29 · answer #5 · answered by hustolemyname 6 · 0⤊ 0⤋
Finding a job Getting used to the gas prices
2016-05-17 09:48:06 · answer #6 · answered by ? 3 · 0⤊ 0⤋
1st question is 2 seconds
2nd question is when c equals 18
2007-05-24 21:00:47 · answer #7 · answered by Anonymous · 0⤊ 1⤋
for the second question, c is 18. Can't understand the first question...
2007-05-24 21:09:20 · answer #8 · answered by Jar 1 · 0⤊ 1⤋