A rocket is launched at t=0 seconds. Its height, in feet, above sea-level, as a function of time,t, is given by
h(t)=-16t^2+112t+480
When does the rocket hit the ground after it is launched?
0 = 16 t^2 - 112 t - 480
0 = t^2 -7 t - 30
0 = (t-10)(t+7)
t = 10
the other root was before it was launched
To find when the rocket hits the ground, we need to determine the value of t when the height, h(t), is equal to zero.
Given that the height function is:
h(t) = -16t^2 + 112t + 480
Setting h(t) to zero, we can solve for t:
0 = -16t^2 + 112t + 480
This is a quadratic equation in the form of Ax^2 + Bx + C = 0, where A = -16, B = 112, and C = 480.
To solve the equation, we can use the quadratic formula:
t = (-B ± √(B^2 - 4AC))/(2A)
Substituting the values from our equation into the quadratic formula, we get:
t = (-(112) ± √((112)^2 - 4(-16)(480))) / (2(-16))
Simplifying further:
t = (-112 ± √(12544 + 30720)) / (-32)
t = (-112 ± √43264) / (-32)
t = (-112 ± 208) / (-32)
We have two possible solutions:
1. t = (-112 + 208) / (-32) = 96 / (-32) = -3
2. t = (-112 - 208) / (-32) = -320 / (-32) = 10
Since time cannot be negative in this context, the rocket hits the ground after it is launched at t = 10 seconds.