Question

How much time passes before the cap hits the ground?

Graduate students throw their graduation caps into the air to convey their hopes and aspirations for a future career. The height, h(t), in feet of the cap in the air t seconds after it is thrown can be modeled by the function:

`h(t) = −16t^2 + 32t+ 4`

Answer 1

Exact answer: `x=1+sqrt5/2 " s"`

Explanation:

To find the time at which the cap hits the ground, the height must be equal to `0`:

So by setting the height equal to `0`, we get:
`-16t^2+32t+4=0`

Solve it as a quadratic:
Factor with GCF:
`-4(4t^2-8t-1)=0`

It is not possible to factor further so let's use the quadratic formula:
`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(8+-sqrt(64-4(4)(-1)))/(2(4))`

`x= (8+-4sqrt(5))/8`

`x=1+-1/2sqrt5 " s"`

Since time can't be negative:
`x=1+sqrt5/2 " s"`

graph{-16x^2+32x+4 [-91.7, 53.6, -21.14, 51.5]}