Algebra 2 problem?

Question

Suppose a baseball is shot up from the ground straight up with an initial velocity of 64 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s = -16t2 + v0t + s0
•16 represents 1/2g, the gravitational pull due to gravity (measured in feet per second2).
•v0 is the initial velocity (how hard do you throw the object, measured in feet per second).
•s0 is the initial distance above ground (in feet). If you are standing on the ground, then s0 = 0.



How long will it take to hit the ground?

Answer 1

For the given values, your function becomes:
s = -16 * t^2 + 64t + 0 (since it's being fired from the ground)

Since s represents distance above the ground, it will be 0 at two different times: When it's fired (t = 0), and when it hits the ground.

Therefore, you want to find the other value of t that makes s 0. Plug into your equation:
0 = -16 * t^2 + 64t

Factoring out a t gives:
0 = t* (-16 * t + 64)

Therefore, either t = 0 (again, the ball is at the ground right when you fire it), or (-16 * t + 64) = 0 (if two numbers multiply out to be 0, one or both must be zero).
-16 * t + 64 = 0

Divide both sides of the equation by -16:
t + -4 = 0

Add 4 to both sides:
t = 4

After 4 seconds, the ball will hit the ground.

Answer 2

First find the equation. Plug in 64 for the initial velocity and 0 for the initial height, because it's shot up from the ground:
-16t^2 + 64t + 0
s = -16t^2 + 64t
s represents the height above the ground, so when it hits the ground, s will equal 0:
s = -16t^2 + 64 t = 0
16t^2 = 64t
16t = 64
t = 4

4 seconds