The height of a fireworks rocket in meters can be approximated by h=-5t^2+30t, where h is the height in meters and t is the time in seconds. Find the time it takes the rocket to reach the growd after it has been launched.
I'm still trying to help my cousin, and this problem isn't making sense to me...help???
2007-01-25 13:45:18 · 6 answers · asked by Adam 2 in Science & Mathematics ➔ Mathematics
First: replace "0" with the h-variable and find the time...
0 = - 5t^2 + 30t
Sec: factor the expression > find the greatest common factor...
0 = - 5t(t - 6)
Third: solve for the two t-variables > set both to equal "0"...
a. - 5t = 0
*Isolate "t" on one side > divide both terms by "-5"...
-5t/-5 = 0/-5
t = 0
b. t - 6 = 0
t - 6 + 6 = 0 + 6
t = 6
*You can only use 6 secs.
2007-01-25 17:02:19 · answer #1 · answered by ♪♥Annie♥♪ 6 · 0⤊ 0⤋
Reaching the ground means h=0.
h= -5t^2 + 30t
h= -5t(t-6)
h=0 at t=0 and t=6.
t=6 is the answer.
2007-01-25 13:54:49 · answer #2 · answered by fcas80 7 · 0⤊ 0⤋
If it is on the ground its height is 0.
So 0 = -5t^2 + 30t
Factor: 5t(-t + 6) = 0
so t = 0 or 6
2007-01-25 13:56:48 · answer #3 · answered by hayharbr 7 · 0⤊ 0⤋
"Reaching the ground" makes the height at that point 0. Plug in 0 for h and solve to t (time).
h = -5t^2 + 30t
0 = -5t^2 + 30t
0 = -5t(t - 6)
Use your zero product property to set each factor to 0.
-5t = 0
t = 0 (starting point)
t - 6 = 0
t = 6
6 seconds.
Hope that helps!
2007-01-25 13:57:32 · answer #4 · answered by teekshi33 4 · 0⤊ 0⤋
Height when it hits the ground is 0
0 = -5t^2 + 30t
0=-5t (t-6)
t= 0 or t = 6
So 6 seconds is solution.
2007-01-25 13:55:29 · answer #5 · answered by leo 6 · 0⤊ 0⤋
first, you have to know how high the rokect goes. then you can work backwards to find the amount of time it took for the roeckt to reach the hieght.
you then double that time since the same time for the rocket to fall back down.
that's how you solve this problem
2007-01-25 14:04:43 · answer #6 · answered by b0b 7h3 l337 2 · 0⤊ 0⤋