Question

A rocket is launched and the quadratic function h(t) = -5t^2+165t relates the height (h) in meters to seconds (t) after launch. When will the rocket be 450 feet above the ground? (Hint...twice!!!) - this is solved by factoring

Answer 1

h(t) = -5t^2+165t

We need to set this equation to 450

450 = -5t^2+165t

Subtract 450 from each side

450-450 = -5t^2+165t-450

0 = -5t^2+165t-450

Divide each side by -5

0 = t^2 - 33t+ 90

Now factor the right hand side

0 = ( t-3) (t-30)

Using the zero product property

t-3 =0 t-30 =0

t =3 t=30

t=3 is when it =450 on the way up

t=30 is when it = 450 on the way down