SOLUTION: A toy rocket is launched from a platform that is 4 feet above the ground. The height, h, in feet, of the rocket t seconds after launch is given by h= -16t^2 + 160t + 4. How many se

Question 396170: A toy rocket is launched from a platform that is 4 feet above the ground. The height, h, in feet, of the rocket t seconds after launch is given by h= -16t^2 + 160t + 4. How many seconds after launch will the rocket be 100 feet above the ground?
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
A toy rocket is launched from a platform that is 4 feet above the ground. The height, h, in feet, of the rocket t seconds after launch is given by h= -16t^2 + 160t + 4. How many seconds after launch will the rocket be 100 feet above the ground?

h = -16t^2 + 160t + 4
plug in 100 feet for h and solve for t
100 = -16t^2 + 160t + 4
(100/-16) = t^2 - 10t - 1/4
-25/4 = t^2 - 10t - 1/4
0 = t^2 - 10t + 24/4
0 = t^2 - 10t + 6
use quadratic formula with a=1, b=-10, c=6:

t = 5 +- sqrt(76)/2
t = 5 +- 2sqrt(19)/2
t = 5 +- sqrt(19)
t = 0.641101 rounded to 6 decimal places seconds
or t = 9.358899 rounded to 6 decimal places seconds
check:
100 = -16t^2 + 160t + 4
100 = -16(0.641101)^2 + 160(0.641101) + 4
100 = -6.576167875216 + 102.57616 + 4
100 = 100, yes
100 = -16(9.358899)^2 + 160(9.358899) + 4
100 = -1401.423847875216 + 1497.42384 + 4
100 = 100, yes
what is highest point and when?
use vertex form y = a(x-h)^2+k, where h,k is vertex
h = -16t^2 + 160t + 4, in this case h is y, and t is x
h = -16(t^2 - 10t) + 4
h = -16(t^2 - 10t + 25) + 4 + 400 (-16 * 25 = - 400)
h = -16(t - 5)^2 + 404, vertex is (5,404)
check with t=5 seconds:
h = -16(5)^2 + 160(5) + 4
h = -16 * 25 + 800 + 4
h = -400 + 800 + 4 = 400 + 4 = 404, yes and highest point is 404 feet at 5 seconds

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